100W LED with a switch
Your PFET has 30 V on it's source.

Depending whether the switch is open or closed, there's either 12 V or 0 V on the gate, so \$V_{gs}\$ is either -18 or -30 V. Either way, the FET will be switched strongly on.

Connect the pull up resistor to 30 V instead of 12 V and it should work.

Note: IRF4905 has a maximum (or rather, minimum) \$V_{gs}\$ of -20 V, so find a way (for example with a voltage divider) to have the switch connect the gate to something like 20 V, instead of to ground, when it is closed.