If it's the only type of transistor in the circuit, the translation is straightforward; build the circuit as designed, reverse the power connections and any other polarised components (diodes, electrolytic caps).
If you need one PNP in a mostly NPN circuit, there is no general solution.
There may be solutions, depending on the configuration of the PNP stage.
For example, if the PNP transistor was being used as an emitter follower, and you have the headroom, you may be able to use an NPN in common emitter, with Rc=Re so that its gain is (approx) 1.
If the PNP transistor was in a complementary power output stage but you can only find low power PNP transistors, I remember seeing an arrangement using a PNP driver transistor and an NPN power transistor to "replace" the non-existent PNP power transistor. Peter Walker did this around 1970 for the Quad 303 power amp (I believe 3 transistors were involved) when there was no PNP version of the famous 2N3055.
And there may be other such substitutions.
Remove R5 and you will have what you describe. The configuration of Q5 is called common collector or emitter follower. Essentially, the voltage at the emitter is the voltage at the base minus 0.6V, but the emitter current can be much more than the base current, because the gain of the transistor will draw more current from the collector. Thus, it's a current amplifier.
Remember, the base-emitter junction is a diode. So, the emitter will be about 0.6V below the base if you forward bias it. With R5 removed, you can pull the emitter up to \$V_{cc} - 0.6V\$. With R5 present, you won't get it as high, since some voltage will be dropped when current flows in R5.
Since there are things that will limit the current in the emitter leg of Q5, you don't need R5 to limit the base current, which isn't true of Q2 or Q4, which have their emitters shorted to ground, or Q1, with its emitter shorted to \$V_{cc}\$.
See Why would one drive LEDs with a common emitter?
There isn't much difference in performance. In circuit 1, the anode of D1 will be at \$V_{cc} - 0.2V\$, whereas in circuit 2, it will be at \$V_{cc} - 0.6V\$, so the LED current is a bit higher in circuit 1, assuming R1 and R4 are the same value.
Circuit 2 has the advantage that the base current goes towards powering the LED, but since the base current is small, this isn't a big effect.
The last subtle difference is that in circuit 1, Q1 enters saturation, which will charge the base-emitter capacitance. When you then turn it off, this capacitance has to discharge before Q1 really goes off, adding a bit of delay from when your MCU output goes low to when the diode gets switched off by Q1. Q5 never enters saturation, because the emitter voltage is brought up to just the point where the transistor enters saturation, but not more. So, no turn-off delay. The delay is very short, and probably not significant until you are switching at least 50kHz.
Best Answer
Any amplifier that can be made with an NPN BJT can also be made with a PNP. Whether its inverting or not really depends on how the output is interpreted. More on that later.
To convert this NPN common-emitter amplifier to a PNP common-emitter amplifier, just mirror the entire thing except for the supply voltages top-for-bottom:
Equivalently, you can just swap the NPN transistor for a PNP transistor, and replace +Vcc with -Vcc. However, the convention is to draw schematics with higher voltages at the top, so the resulting schematic would look a bit funny.
Remember that voltages are relative. The only important thing with a PNP common emitter amplifier is that the emitter is at a higher voltage than the other terminals, the base is approximately 0.6V lower than the emitter, and the collector will be lower than the emitter, with how much lower controlled by the base current and load.
If we call the highest voltage in the circuit "ground", then we can have negative voltages. Or, we can call the lowest voltage "ground" and have positive voltages. We can even pick a voltage in the middle and have both. Or, we can ignore ground altogether and talk about the voltage "drop" or "across" a component or between any two points in the circuit.
This is the same circuit, just with a different notion of "ground", which is completely irrelevant to the operation of the circuit, only our discussion of it:
Really, the terminal labeled "output" is a voltage somewhere between the other two terminals. Is it inverting? Well, are we considering the signal to be the output relative to the higher voltage at the top, or the lower voltage at the bottom?
Let's get rid of the names which the electricity doesn't know about, and draw the whole circuit, with a power supply and all:
There is no "ground" and there is no "Vcc"; there's just a battery with two terminals, one with a higher potential than the other. We can call them what we like; the circuit doesn't care, as long as there's that voltage difference there.
There is also no "output" terminal, but instead we have two voltage differences, either of which could be considered the "output": \$V_a\$ and \$V_b\$.
When the input \$V_{in}\$ is low, this forward-biases the transistors base-emitter junction more, turning it on more, making it effectively look like a smaller resistor and allowing more current in \$R_L\$. By Ohm's law, if the current over a resistor increases, so too will the voltage across it. Or, you can think of \$R_L\$ and the transistor as making a voltage divider. Either way, you can see than when \$V_{in}\$ goes down, \$V_b\$ goes up.
\$V_a + V_b\$ must be equal to the battery voltage, since they are in parallel with the battery. So if \$V_b\$ is going down, \$V_a\$ must be going up to make up the difference.
So is it inverting? I can't say! Is \$V_a\$ or \$V_b\$ the output?
What if \$V_{in}\$ is connected between the + side of the battery and C1, instead of the - side of the battery and C1, as it is now? Then is this an inverting amplifier, or not?