Let's assume the incircle is tangent to $AB$ at $F$, then we have that $\angle FAD = \angle IAQ$ and $\angle AFD = \frac{180^\circ - \angle B}{2} = \angle AIQ$ where the first equality is from the statement and the other are well-known identities. It follows that $\triangle AFD$ and $\triangle AIQ$ are similar to each other and hence $\triangle ADQ$ and $\triangle AFI$ are also similar to each other. In particular $\angle ADQ = \angle AFI = 90^\circ$ and moreover, by a similar argument, we get the analogous equality $\angle ADP = 90^\circ$. Finally, we observe that $\angle PAD = \angle CAI = \angle BAI = \angle QAD$ and hence $AD$ is the angle bisector and also the altitude that corresponds to vertex $A$ in triangle $\triangle PAQ$. Thus $AP = AQ$.