Electrical – Circuit works with BC557 PNP transistor, but not TIP32C

Page 2 of this datasheet tells the story.

Under absolute maximum ratings it states that the maximum reverse voltage of base and emitter is 5V. You have exceeded this by using 12V and although the base-emitter region is probably still intact (due to the 100k resistors in your design) you can't expect leakage current to be insignificant.

Don't expect simulation models to go beyond (or even get near) real-world issues due to absolute maximum ratings - models expect you to obey rudimentary principles and this is one of them.

Try a 5V control signal.

Also, when power is off you won't be able to conduct much signal thru from emitter to collector because you are relying on the base-emitter junction to be foward biased and this requires at least +0.7 volts on the emitter to do so.

You should put a resistor in series with each LED and eliminate R18. If you are just switching LEDs it's perhaps wasteful to construct a high-side switch- if you can switch the low side it only requires one NPN transistor per LED.

The high-side switch you show (top diagram) will work, however you can only switch a fairly small current due to the 10K base resistor. At 12V you'll get about 1mA base current so most transistor will be well saturated for up to ~20mA load current, but if you want to switch 100mA reliably you should reduce that resistor in most cases.

Also, it's good practice to add a resistor from base to emitter on the PNP. The reason is that the leakage in the NPN can be amplified by the PNP and result in excessive current at the output (particularly at high temperatures). Something like 20K to 100K will work fine (it's not critical). That said, you can guess that the gain of the PNP will be low at low base current and the circuit will typically work fine at moderate temperatures.