1-5V to 4-20mA current loop transmitter
If the voltage across R3 is not equal the op-amp drives the MOSFET gate in the correct direction to increase or decrease the current through the load (assumed to be connected between R6 and ground) so that the voltages are equalized.
If the load current is too low then the voltage at (-) input will be too high (voltage across R3 is too low) and the op-amp output is driven low, which (being a P-channel MOSFET with source connected to +24) will cause the MOSFET to turn on more, and drive the voltages closer together.
The voltage across R2 is driven by the input voltage and resulting drain current from M2 (through the LH op-amp). The voltage across R3 is driven by the RH op-amp  to equalize the voltages, via changing the load current. Since there is no gate current to speak of, the current through R3 is about exactly equal to the load current.
If the load resistance is too high (say it is open) then the MOSFET will have -24V Vgs but no drain current to speak of, so the voltages cannot equalize.
Note that this will exceed the absolute maximum rating of most MOSFETs so maybe this design is not ideal. I don't see a datasheet for an IRF9Z245, but the IRF9Z24S (maybe a typo) is only rated for -20V absolute maximum.