The circuit is good for protecting the diode. I don't know of a better one.
You don't have to worry about protecting the diode from transients. If a large current ( < 1A) flows through the diode for a microsecond, it won't hurt the diode. You can get away with a regular zener diode instead of a TVS.
You might consider using a lower zener (TVS) voltage, say 5V. That will cause less variation of diode current between 12V and 24V input voltage.
Optoisolators tend to age poorly, and the CTR will decline over time (or at least they used to, maybe that's been improved). So it's a good idea to drive the diode with more than the minimum current to be sure that over time the input drive will continue to drive the output. 10mA is probably a good number for an opto LED spec'd at 5mA minimum.
Despite several requests for detailed requirements, not nearly enough has been forthcoming, and it has been requested "What I am looking for is feedback regarding the circuit itself, addition of decoupling capacitors, impedance matching problems, timing problems etc.. not the models themselves". Well, I'll do my best. What requirements are known are
-ESD protection provided by a one channel IC
-low pass filter to suppress noise above 4.5Khz
-pmos to protect against reverse polarity
-crowbar for overvoltage protection
-current sensor which is connected to a uC ( ideally I would like the op amp to work with the PSU IN in the range 3.3v to 16v)
The maximum voltage is 16V and the minimum votlage is 3.3V.
So let's take it step by step.
ESD protection. Since the schematic parts are not necessarily what will be used, this is a bit tricky, but let's stick with the schematic, and specify a TPD1E10B09. Data sheet is here http://www.ti.com/lit/ds/symlink/tpd1e10b09.pdf. The most obvious problem is that it acts as a bilateral 12 volt zener, as shown in Fig. 5. Since this is an 0402 package, it will be destroyed by any sort of input above 12 volts. 16 volts is not an option.
Low-pass filter. This is a pi filter with two 47 uF caps and a 27 uH inductor (inductor resistance unspecified). Since the PSU input can be assumed to be very low resistance (it's a power supply output), the first capacitor is essentially irrelevant. The resulting L filter does indeed roll off after about 4.5 kHz. However, its response looks like 
and is not what I would recommend to get rid of noise. That's a gain of more than 1000 at 4.2 kHz. The peak can be eliminated by adding a 1 ohm resistor in series with the inductor.
Crowbar. This uses a 1N750 zener and a C233 SCR. The zener has a knee voltage in the range of 4.5 to 4.9 volts, and the SCR is guaranteed to fire with a gate voltage of 1.5 volts and 9 mA. The result is a crowbar voltage of as much as 6.4 volts. If this is adequate to protect the pcb, fine. However, without knowing the characteristics of the power supply providing the power into the PSU IN, it is not clear what will happen to the 27 uH inductor if the crowbar activates. Smoke and flame is a real possibility. ETA - This is wrong. The fuse will protect the inductor. Sorry.
PMOS reverse bias protection. A p-type MOSFET is used. Again, there is no indication of exactly which MOSFET will actually be used, and the schematic unit is clearly unsuitable, so let's assume something like an IRF9530. With a 3.3 ohm load (allowing 1 amp at 3.3 volts output, the circuit was simlulated for a 0-5 volt input:
Green is input, brown is output. A 3.3 volt input will only produce a 1.4 volt output, so maybe some rethinking is in order.
Current sensor. It was later specified that the current monitor output should be in the range of 0 - 3.3 volts, but the maximum current level was not specified. The schematic circuit shows a voltage amplifier with a gain of 7. 3.3 volts / 7 equals .47 volts. The resistor which is obviously intended to be the sense resistor has a value of .47 ohms, leading to the conclusion that the desired current sense range is 0 - 1 amp. The schematic circuit simply does not measure current, so the following is offered
simulate this circuit – Schematic created using CircuitLab
This is a standard difference amplifier with a gain of 7, in line with the original. The op amp is powered by the PSU input, with a 10 ohm / 10 uF power filter. While an appropriate op amp can, I'm sure, be procured, there are a few things to remember. The opamp does not need to be rail-to-rail at the input. The power supply, due to the ESD clamp, will not get above about 12 volts.
However, I cannot recommend this circuit. The 0.47 ohm sense resistor will drop .47 volts when the current is one amp. Adding this to the drop in the PMOS section, the 1 volt drop produced by the damping resistor in the low-pass filter, and the unspecified drop in the fuse, a full-scale current will produce an awful lot of voltage drop. (Speaking of fuses, a Littlefuse series 208 2AG fuse rated at 1 amp has a cold resistance of 0.103 ohms.) A better approach would be to use a smaller sense resistor and a larger gain in the difference amplifier.
To sum it up, I am dubious about every part of the circuit.
Best Answer
This is all you need:
simulate this circuit – Schematic created using CircuitLab
this also eliminates the problem that a high voltage at the input will lift-up your local VCC.
When input < -0.6 V D1 will conduct and limit GPIO to -0.6 V, R2 limits the current, your GPIO input will be able to handle this (it also has input protection diodes !)
When input > -0.6 V but < 3.9 V D1 does nothing, GPIO also happy
When input > 3.9 V D1 will conduct, R2 will limit any current, GPIO input will be happy.
Someone complained this would not work but was too lazy to explain why but I figured it out myself:
Apparently I overlooked that 3.9 V zenerdiodes leak a lot in reverse so I lowered R1 to 10 kohm. If that still doesn't fix it than you could replace the zenerdiode with 4 to 5 standard diodes in series, see 2nd schematic.