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Since $ABQ$ and $BCQ$ are equilateral, the angles $ABP$ and $CBQ$ are both $60^\circ$. So $$\angle{PBQ} = 360^\circ-90^\circ-60^\circ-60^\circ=150^\circ$$
PBQ is isosceles, so the angles $BPQ$ and $PQB$ are equal. So
$$2 \times \angle{PQB} = 180^\circ- 150^\circ = 30^\circ$$ So $$\angle{PQB} = 15^\circ$$
This problem is taken from the UKMT Mathematical Challenges.
You can find more short problems, arranged by curriculum topic, in our short problems collection.