How to calculate capacitor value to reduce PWM load ripple?
Without knowing much about the power supply how your power source was designed, just assume the worst case.  Which is that it doesn't have significant output capacitance to absorb the ripple.  Then calculate how much capacitance you need to achieve a certain amount of voltage ripple.  Any capacitance your power source has will just make that estimate more conservative.

The voltage ripple on a capacitor is calculate as..
dv = I * dt / C

dv = the peak to peak voltage ripple in Volts.
I = the current which is being switched in Amps.
dt = the time that the capacitor is supplying current in seconds.
C = the capacitor value in Farads.

In your case of a 500Hz signal, at 8A, we have dt = 1ms, and I = 8A.  Lets say we want the voltage ripple (dv) to be less than 0.1V then...

C = I * dt/dv = 8A * 1ms / 0.1V = 80mF

Some power supplies become unstable when large capacitances are added to their output.  Therefore you may want to add a small value resistor or inductor between your supply output and the capacitor.  Something on the order of 0.1 ohms, 10W is about right.