Purpose of a capacitor between op amp input and output
It is to stabilize the 120mA current sink (the function of this circuit block). The 10K resistor (which is too high, IMO) R13 introduces a pole in the feedback loop, reducing the phase margin, in conjuction with the gate capacitance. 

The 1uF and R17 provide a faster AC feedback that negates that, at the expense of rather sluggish response, particularly so because of the values chosen. 

You could decrease R13 to something like 100 ohms and C12 to 10nF or even 1nF, probably, with improved AC performance. 



As to how it works, the R11/R14 voltage divider from +5 yields 238mV at the non-inverting input to the op-amp and the op-amp drives the MOSFET gate to match that voltage at the inverting input, which requires 120mA to flow through the source in order to drop 238mV across the 2 ohm resistor R19. 

The simple way to look at it would be to eliminate C12 (open) and R13/R17 (short) and it looks like it might work, but it would probably oscillate or oscillate conditionally.