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.*