If $P' \in AP$ such that the perpendicular from $P'$ to $BC$ passes through the $A$-excircle's tangency point on $BC$, then $APQR \sim AP'BC$ by homothety and $BPP'C$ is an isosceles trapezoid, so $\angle{BPC} = \angle{BP'C} = \angle{QPR}$ which implies the desired result. $\square$

Let the incircle hit $BC$ at $D$, $AC$ at $E$ and $AB$ at $F$. Let the tangency point be $D'$. Let the incenter be $I$. Use complex numbers with the incircle as the unit circle and let $D = -i$ (so $ID$ is the complex axis). Let $E = e$ and $F = f$. By the tangency formulas we get $$b = \dfrac{-2if}{-i + f}$$$$c = \dfrac{-2ie}{-i + e}$$and $$a = \dfrac{2ef}{e+f}.$$By parallel lines we have $DD' \perp RQ$ and so $DD'$ is a diameter. Therefore, $D' = i$. By tangency formulas, $$q = \dfrac{2fi}{f+i}$$and $$r = \dfrac{2ei}{e+i}.$$Consider point $P = p$. Since $P$ lies on the complex axis we know $p + \overline{p} = 0$ and $\angle{APO} = 90.$ So, $$\dfrac{a-p}{-p} + \dfrac{\overline{a} - \overline{p}}{-\overline{p}} = 0.$$Replacing $\overline{p} = -p$ and solving gives $$2p = a - \overline{a} = \dfrac{2ef}{f+e} - \dfrac{2}{f+e} \implies p = \dfrac{fe - 1}{f+e}.$$It suffices to show $$\dfrac{(b-p)(r-p)}{(q-p)(c-p)} \in \mathbb{R}.$$This is a bashy check, but it works.

Let $G$ be the touch point of the incircle and $BC$ and $D$ be the antipode of $G$ and $E, F$ be the other two touch points. Let $H=BR\cap CQ\cap EF\cap DG$ (by pascal). We calaim that $PH$ is the bisector of $\angle BPR$ and $\angle CPQ$ and this finishes the problem. Let $J=AP\cap BR$. Notice that $(JH;RB)=(PH;DG)=-1$. Since $\angle HPJ=90^\circ$, so $PH$ bisects $\angle BPR$.