We first show that $Q$ lies on the circumcircle of triangle $ABC$.

Let the perpendicular bisector of $BC$ intersect the circumcircle of $ABC$ at $Q'$. Since $Q'$ bisects the arc $BC$ we have $\angle BAQ' = \angle CAQ'$. Thus $Q'$ lies on the angle bisector of $\angle BAC$. Hence $Q = Q'$.

Let $BC = a$, $AC = b$ and $AB = c$. From the Angle Bisector Theorem we have

$$BP = \frac{ac}{b+c} \quad \text{and} \quad PC = \frac{ab}{b+c}. \qquad (1)$$

The triangles $ABP$ and $AQC$ are similar, thus
$$\frac{c}{AP} = \frac{AQ}{b}. \qquad (2)$$

On the other hand, computing the power of point $P$ with respect to the circumcircle of $ABC$ we get
$$BP \cdot PC = AP \cdot PQ. \qquad (3)$$

Combining (1), (2) and (3) yields
$$\frac{PQ}{AQ} = \frac{BP \cdot PC}{bc} = \frac{ac}{b+c} \cdot \frac{ab}{b+c} \cdot \frac{1}{bc} = \left(\frac{a}{b+c}\right)^2 = \left(\frac{BC}{AB+AC}\right)^2$$

as required.