I don't think the resistor in the diode path has anything to do with biasing (which can also be seen in your simulations). It has to do with other factors.
Connecting a supply net directly to the gate means that if there is a large spike on the supply, then the gate will see that large spike and can cause the oxide to breakdown. Putting that resistor there means you have a low pass filter from your supply to the gate and any spike on your supply voltage will be low pass filtered by the resistor right there.
Also, directly connecting the gate to the supply means you are connecting a capacitance to the supply net. This in combination with the bondpad inductance is a perfect oscillator. By placing that resistor in series with your gate, you create a poor quality capacitor which makes sure you kill any oscillations. People do use large decoupling capacitors on chip but that is well controlled as the supply networks are simulated separately to rule out any supply oscillations. But doing such a simulation with your full chip and all analog blocks in it is extremely time consuming. To rule out any inadvertent effects, designers often place such resistors in the feedback path.