Good problem, kinda surprised to see that it's related to mixtilinear circles. Let the circumcircle of triangle $ABP$ be $\gamma$. Let $F$ be the intersection point of segment $AE$ and $\Omega$, and $X$ be the intersection point of line $BF$ and $\gamma$. It is obvious that $AB=AC$. $\angle XPA=\angle XBA=\angle FBA=\angle FEB=\angle AEB=\angle APB$. Therefore $AB=AX$. $\therefore AB=AC=AX$, and $\angle APB=\angle APX$. By the Mansion's theorem in $\triangle BPX$, $C$ is the incenter of $\triangle BPX$. Let $\angle CAB=\theta$. $\angle CEB=\angle CBA = 90 - \frac{1}{2} \theta$. Also $\angle CEP=\angle BEP - \angle BEC=180-\theta - (90 - \frac{1}{2} \theta )=90 - \frac{1}{2} \theta$. $\therefore CE$ is the bisector of $\angle PEB$. Let the $X$-mixtilinear circle of $\triangle XBP$ be $\xi$. Let the tangent point of $\xi$ and $\gamma$ be $E'$. Let the midpoint of the arc $\widehat{BXP}$ be $M$. Since it is well known that $(E',C,M)$ are collinear, (see the second lemma in https://www.awesomemath.org/wp-pdf-files/math-reflections/mr-2020-02/mr_2_2020_mixtilinear.pdf) $E'=E$. Therefore $E$ is the tangent point of the mixtilinear circle $\xi$ and the circumcircle $\gamma$. Note that $\angle PEQ=2\angle APB=\angle BPX$. Therefore $\angle PEQ=\angle BPX=\angle BEX$. So $EQ$ and $EX$ are isogonal lines at $\angle PEB$. Meanwhile, let $Q'$ be the tangent point of $BP$ and the incircle of $BPX$. It is well known that $EQ'$ and $EX$ are isogonal lines at $\angle PEB$. (See the 17th lemma in https://www.awesomemath.org/wp-pdf-files/math-reflections/mr-2020-02/mr_2_2020_mixtilinear.pdf ). Since the intersection point of $BP$ and the isogonal line of $EX$ at $\angle PEB$ is unique, $Q'=Q$. Therefore $BP$ and $CQ$ are perpendicular.

Interesting Problem