How does the back-power protection of the pi function?
1.) whith positive voltage current will flow through Q2A and as it is a current mirror, it will also open Q2B. Together with the parasitic diode of the mosfet, it will rise the voltage at R24. This will open the mosfet -&gt; normal operation.
In case of under/reverse voltage, the circuit will close the mosfet (current mirror has a high gain, so it will discharge the gate quite fast) and the parasitic diode is also in blocking direction
2.) datasheet of the mosfet says no:


Example: reverse polarity protection + voltage drop buffer

One complete cycle

The reverse polarity protection working: the voltage drop buffer is discharged by the load, the charge is not flowing back to the source.
The load is backed for almost 50ms (goal is 10ms, industrial standard)

so after 10ms the load still has ~4V

The simulation can be downloaded here.