If you have an open-collector or open-drain type logic output, you can drive a PNP with it, as long as the device is rated to take your supply voltage. In this configuration, use a pull-up resistor from the supply to the base of the PNP to ensure that the PNP defaults to the OFF state. Now tie the base of the PNP to the open-collector logic output through a second resistor, the value of which should pull just enough current from the base of the PNP to drive it to saturation.
The values aren't too hard to work out. The pull-up usually wants to be something large-ish, like 50K or 100K. The logic resistor will see the supply voltage, less the saturation E-B drop of the PNP, less the Vce sat of the logic output. Basically, the supply voltage minus about a volt. From there, knowing the necessary base current to the PNP will let you work out the exact value. (Be sure that to saturate, the necessary PNP base current isn't actually greater than the current-sink capacity of the logic pin!)
If you do not have an OC/OD output, you can buffer a regular output signal with a small NPN transistor, like a 2n2222 or 2n3904 etc. A typical TTL output, driving the base of a small NPN through about 1K to 2.2K will give you the sort of open collector (the collector of the NPN) that I've assumed above.
If it's the only type of transistor in the circuit, the translation is straightforward; build the circuit as designed, reverse the power connections and any other polarised components (diodes, electrolytic caps).
If you need one PNP in a mostly NPN circuit, there is no general solution.
There may be solutions, depending on the configuration of the PNP stage.
For example, if the PNP transistor was being used as an emitter follower, and you have the headroom, you may be able to use an NPN in common emitter, with Rc=Re so that its gain is (approx) 1.
If the PNP transistor was in a complementary power output stage but you can only find low power PNP transistors, I remember seeing an arrangement using a PNP driver transistor and an NPN power transistor to "replace" the non-existent PNP power transistor. Peter Walker did this around 1970 for the Quad 303 power amp (I believe 3 transistors were involved) when there was no PNP version of the famous 2N3055.
And there may be other such substitutions.
Best Answer
You are trying to create a high-side switch.
simulate this circuit – Schematic created using CircuitLab
Figure 1. (a) NPN open collector switch. (b) A failed attempt to make a PNP version.
I can tell you that the above configuration is not correct and can not be used safely.
The low side switch is very simple as shown in Figure 1a. The low-voltage logic just has to feed sufficient current into the NPN transistor to turn it fully on.
There is a temptation to think that we could do the same trick with an PNP transistor as shown in Figure 1b. The problem is that the emitter-base junction is always forward biased. This will apply the 12 V to the chip output and destroy it or, if there are protection diodes on the output, the current will flow through the protection diodes into the micro-controller supply (shown as 5 V in this case). The effect of this current flow is to turn on Q3 and the load can not be switched off.
simulate this circuit
Figure 2. A high-side switch.