Hi,
The NPN is backwards, and the 1k R10 should be on the PIC enable pin side to limit input current, surely? If so, it could be a larger value, say 10k. The emitter should be facing the PIC input, not the regulator (see how you have a diode's cathode facing the output it uses?). It might have been better to use a PNP in the signal path instead, to get the most out of the 5V - 1k signal, but now that would mean using two transistors and it would "be wasteful", comparatively speaking: a PNP on/off switch for the enable path and an NPN to turn the PNP on/off.
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Hi again,
You could be better served by sticking to your original circuit and just put an e.g. 10k or even better still 100k from the PNP base to ground and make R4 100k or better yet 1M and R1 10k. There's no power available to the PNP emitter unless the regulator is on, right?
You'll never get 5V with things in the way. A resistor drops a finite voltage based on current passing through it, a PNP or an NPN drops Vce voltage based on its internal resistance... If the PIC doesn't need over e.g. 4.5V, at a guess, the PNP version should be okay. Maybe forget the BJT switch completely if on reflection of what's happening there it is superfluous to the 5V being presented or not to the PIC enable pin. Hope I've understood what you're doing.