S = 5v G = 4.29v D = 5v
How do you get G = 4.3 V? The gate is connected directly to ground, and the gate-source diode is reverse biased so it plays no role. The gate will be at 0 V here. (Unless there's something else connected to the REVERSE_GATE node that you haven't shown in the schematics you shared)
if a P-channel mosfet turns on when it sees 0v at the gate
A P-channel (enhancement mode) MOSFET turns on when it sees a gate-source voltage more negative than its threshold voltage.
The voltage relative to the circuit reference (ground) is irrelevant.
why even have that mosfet if it's always going to conduct having the gate at ground? What's the benefit
This is a reverse protection circuit.
Note that current will normally flow from drain to source, rather than the usual direction for PMOS of source to drain. This gets the body diode oriented the right way for a reverse protection circuit.
Now if you accidentally reverse the power supply, you'd be connecting the drain to -5 V. Then Vgs will be positive and the FET will not conduct; also the body diode will be reverse biased and will not conduct. This prevents any power being delivered to the rest of the circuit, so it won't be damaged.