Help with a PNP Switch
Thanks for the question and for the schematic :-)

With the SPST switch open, the resistive divider settles the base at about 120 mV, which is below the opening voltage of a BJT B-E junction. So the transistor does not conduct from C to E. R2 is too high to have any meaningful effect, you could as well leave it out.

Note that you do not have a resistor in series with the SPST switch. If you close the SPST switch = make it conduct, the circuit between your 12V PSU, the switch and the BJT's B-E junction has no defined current limit. If your 12V PSU is strong enough, it will blow the base connection out of the transistor, or fry the chip, or both.

I suggest that you put a resistor in series with the SPST switch. Check how much current your transistor can sustain. In a randomly chosen datasheet for the commodity 2N4033, I cannot see Ibe(max) specified, but Ic(max) is 1A, so you can probably consider that. Actually much less should be sufficient. To allow for Ic=1A, considering a conservative Hfe=50, you could ballpark that you need to limit the base current to 20 mA. At about 11 V available differential voltage (the B-E junction will clamp at about 0.6 V) the resistor should be 11 / 0.02 = 510 Ohms (a particular standardized value in the series).

EDIT:

See Spehro Pefhany's answer.