If you want to switch from the high side (like delivering a specific voltage to a pin), use a PNP transistor.
If you want to switch from the low side (like turning on an LED in the collector circuit), use an NPN transistor.
But, even though the PNP will work as a switch, a uC (or whatever) can't stand the required base voltage to turn it off. That's why your original circuit would work well. The NPN was controlling the base of a high side switch. When the NPN was off, the 10k resistor held the PNP base high so no current would flow.
There are a couple of transistor relationships that are worth knowing: There must be about 0.7V difference between the base and emitter to get the device into saturation. For the NPN case, the base needs to be higher than the emitter. For the PNP case, the base needs to be lower than the emitter.
The other relationship is that, at best, you can get VCE Sat down to 0.2V - that is, the voltage between the collector and emitter when the transistor is saturated will be about 0.2V. You may not be able to get that far into saturation. The 2N3904 datasheeet suggests it may be difficult to get below 1.0V.
I suppose we can add one more relationship: The base current needs to be 1/10th of the collector current to get the transistor into saturation. If you need 100 mA of collector current, you need to design in about 10 mA of base current. This is just a 'rule of thumb' and varies considerably by device. But it's a good place to start!
In your original circuit, the 4.3k resistor set the PNP base current. LTSpice shows the output, given a 390 Ohm load as 11.8V at around 30 mA. The PNP has only 0.2V between the collector and emitter which is deeply saturated. That's a good thing.
Notice how the 390 Ohm load resistor and the 4.3k Ohm base resistor have a nearly 10:1 relationship.
So, you original circuit, with the real transistors simulates pretty well. LTSpice .asc file attached