Why wouldn't it?
Since MOSFETs have no gate current and thus no D3 current, the drop across D3 is 0V and the MOSFET's gate (not base) voltage is the same as the input voltage.
The load has nothing to do with the gate voltage.
The 2N6660 not a BJT and hence there is no base. The 2N6660 is a FET and there is a gate at pin-2.
The input resistance is very high. In other words, the gate current is very low, less than 1 μA.
The voltage drop across the diode D3 at <1 μA is almost zero. Hence the voltage at the gate is 5 V.