From what I can tell with a little bit of hand-wavy analysis and backed up by information in the datasheet, is that it all comes down to wavelength - in fact a since deleted answer touched on that, but I think with the wrong conclusion.
TL;DR
The 200mm cable is closer to an integer number of whole wavelengths at the frequencies of interest than the 150mm cable, meaning the impedance of the antenna as seen by the source is a closer match and so less signal is reflected back from the antenna and so more of it is radiated.
Firstly, lets understand the wavelengths we are talking about. The specifications in that datasheet are for the \$824\mathrm{MHz}\$ to \$960\mathrm{MHz}\$ region, and the \$1710\mathrm{MHz}\$ to \$2100\mathrm{MHz}\$ region. For simplicity, lets say those are roughly \$1\mathrm{GHz}\$ and \$2\mathrm{GHz}\$. In fact it is a little lower than this, but never the less to keep nice round numbers, lets go for that.
Now lets also have a quick look at the cables. You have one which is 200mm long, and a second which is 150mm long. Now we all know that for an EM signal in free space, the frequency and wavelength relationship is as follows:
$$c = f\lambda$$
Where \$c\$ is the speed of light, \$f\$ is the frequency, and \$\lambda\$ is the wavelength.
However the critical thing here is that we are not talking about free space, but rather a cable, and this does actually change things a little. The equation is really:
$$v = \nu_p c = f\lambda$$
Where \$v\$ is the actual propagation velocity, and \$\nu_p\$ is the velocity factor. Without going into detail, for coax cable, \$\nu_p \approx 0.67\$, meaning we can say:
$$f = \frac{2\times10^8}{\lambda}$$
Now for our two pieces of coax, one is 200mm, and the other is 150mm. These correspond to \$1\mathrm{GHz}\$ and \$1.33\mathrm{GHz}\$ respectively. In fact if we are talking about integer multiples of wavelengths (more later), then we also can find that they could correspond to \$2\mathrm{GHz}\$ and \$2.66\mathrm{GHz}\$ respectively as well (or other multiples).
Now for the interesting bit - notice how the frequencies for an integer number of wavelengths in the 200mm cable match the frequencies specified in the datasheet (again roughly)? I think this is the key to answering your question
So why are integer numbers of wavelengths so important? Well, it all comes down to impedance transformation. When you add a transmission line, the signal within the cable is not actually electrical anymore, but now an EM wave contained within the cable. To cut a long story short (you can google the rest), if you have an impedance at one point in the cable, it will actually appear to change impedance as you move along it. This impedance change is cyclical, and at every integer wavelength, the impedance will appear unchanged.
Now what does this mean for you? Well, you have an antenna which presents a specific impedance. In order for the cable to be completely invisible, you need the impedance at the input to be the same as the impedance at the output. Ignoring resistive losses in the cable (*), from the above we can see that every integer wavelength of cable, the load will appear to the source completely unchanged.
From the above calculations, we can see that for the frequencies of interest, the 200mm cable is almost an integer number of wavelengths. As a result, the cable will appear as completely transparent (*).
Whereas for the 150mm cable, it doesn't equal an integer number of wavelengths at any of the frequencies of interest. As a result, the impedance of the antenna will be transformed to a different value, meaning it no longer matches what the source expects.
Impedance mismatches in transmission lines are bad. It can be imagined like a pane of glass. If you shine a laser on the glass at a slight angle, you will see several points appearing. Each one is caused by some of the light being reflected at the impedance mismatch between the air and glass. The same is true for transmission lines. An impedance mismatch causes some of the signal to be reflected which in turn means less will signal power will reach the antenna and be radiated.
It should then be clear as to why the 200mm cable performs better - it constitutes a closer match to an integer number of wavelengths than the 150mm cable at the frequencies of interest, which means there will be less of an impedance mismatch and so more power delivered to the antenna. This is all despite the fact that the 200mm cable has 33% higher resistive loss than the 150mm cable - the impedance mismatch losses likely dwarf the resistive losses.
The frequency ranges are not exactly \$1\mathrm{GHz}\$ or \$2\mathrm{GHz}\$, so the 200mm cable is also not perfectly matched, but it is a much closer match than the 150mm.
You may be wondering, is any of this in the datasheet you posted. Actually it is if you understand what a Smith Chart represents. Page 5 top right shows how a \$50\Omega\$ load impedance is changed by the cable at different frequencies - notice how the line appears to spiral around the central (\$50\Omega\$) point on the Smith chart? This is the impedance changing as viewed by the source.
The above is supported by the diagram top-left of Page 5. Notice how at 1GHz and 2GHz there is are massive dips in the S11 trace. S11 is the reflected signal caused by impedance mismatches and shows signal which was not delivered to the antenna - the lower the number, the less the signal has been reflected and hence the better the impedance match.
(*) The cable actually has some resistive loss which doesn't constitute an impedance mismatch, but rather simply attenuate the signal.
Best Answer
A good reason to put such a loop on the end of the monopole antenna is to reduce the likelihood of someone being stabbed by accidentally falling on or tripping into the antenna. Under all ordinary circumstances, it would have no discernible effect on the performance of the antenna when compared to the same antenna with the small loop uncurled.