Let $I, I_a$ be the incenter and $A$-excenter of triangle $AQR$. To prove the claim of the problem, it suffices to show that $\widehat{BIC} = 90^\circ + \frac{1}{2}\widehat{BAC}$.

We have
$$\begin{aligned} \widehat{ARC} &= \widehat{PCB} - \widehat{RPC} = \widehat{PCA} + \widehat{ACB} - \widehat{ABC} = \widehat{PCA} \\ \implies AC^2 &= AP \cdot AR. \end{aligned}$$
On the other hand, note that points $I, R, I_a, Q$ lie on circle with diameter $II_a$, and
$$\widehat{QIR} = 90^\circ + \frac{1}{2} \widehat{QAR} = \widehat{QPR}.$$
Therefore $PIQR$ is a cyclic quadrilateral and $P$ also lies on circle with diameter $II_a$. Now we have
$$AI \cdot AI_a = AP \cdot AR = AC^2.$$
Let $D$ be the intersection point of $AI, QR$, and let $B'$ be the second intersection point of the incircle of triangle $ICI_a$ with $BC$. We know that this circle is tangent to $AC$, therefore
$$-1 = C(AD, II_a) = C(CB', II_a).$$
So $AB'$ is also tangent to the incircle of triangle $ICI_a$ which implies $AB' = AC$, therefore $B'$ and $B$ are coincident. Finally we conclude that
$$\widehat{BIC} = 180^\circ - \widehat{ABC} = 90^\circ + \frac{1}{2} \widehat{BAC}.$$
Hence the claim of the problem.