Sinking current from 9V into a PIC pin.

[Charlie]

Any base-emitter pair can only be a diode drop apart if the device is working. So, if the emitter is at 9V then the base is at 8.4 V or so. This means that either the PIC RC5 pin would need to be above 8.4V, or tri-stated to an open circuit to turn the transistor off. At 5V or 0V, the transistor is on. So you must be changing the tri-state, not the value (0,1) of the pin.
The problem with this is that all of the pins are briefly exposed to 9V, albiet through current limiters. Maybe the device has a sense of humour about this, maybe not. If you are making a hobby toy for yourself - have at it. If you are making a product, do the support folks a favour and do a little reading about transistors.
You have not shown us your code, or which device you are using so we are guessing here, but rest assured the concerns are valid. Just because it *appears* to work, does not mean it still will next week.

[HankMcSpank]

This is the point where Hank asks for a large dose of humble pie...obviously at some stage in my frenetic breadboarding activity (moving wires around all over the shop), I've actually got 5V going to the PNP emitter (not the 9V I thought was on there).

Just to prove the point...I put 9V on the emitter...the PIC lived, but the LEDS didn't turn off when the duty cycle went to 255.

You're all vindicated! (not that I ever doubted you!)

Charlie...this is just a hobby.