$PR,PQ$ cut $BC$ at $S,T$ and $BR,CQ$ cut $AP$ at $D,E,$ respectively. Line passing through $A \equiv QB \cap RC$ and $P_{\infty} \equiv QR \cap BC$ (lying at infinity) is the polar of $U \equiv BR \cap CQ$ WRT $(I).$ Then if $(I)$ touches $BC$ at $M$ $\Longrightarrow$ $U \in IM.$ We have $P(B,R,U,D)=P(B,S,M,P_{\infty})=-1$ $\Longrightarrow$ $M$ is the midpoint of $BS.$ Similarly, we have $P(C,Q,U,E)=P(C,T,M,P_{\infty})=-1$ $\Longrightarrow$ $M$ is the midpoint of $CT.$ As a result, triangles $\triangle PBS$ and $\triangle PCT$ are both isosceles $\Longrightarrow$ $\angle CPR= \angle QPB.$

Let $\{D\}\equiv \omega \cap BC$, $\{E\}\equiv \omega \cap QR$, $\{F\}\equiv AD \cap QR$, $\{S\}\equiv PR \cap AB$; it is easy to see that, due to the fact that the triangles are homothetic at $A$, $F$ is the contact of the incircle of the triangle $AQR$ with $QR$ and $ER = QF$ $(1)$. From $BC \parallel QR$ we get $\frac{AQ}{AB}=\frac{QF}{BD}=\frac{ER}{DS}$ and, with $(1)$ we get $BD=DS$, so $BP$ is the reflection of $PR$ in $PI$. Similarly $PQ$ and $PC$, so, by difference, we get the required equality. Best regards, sunken rock

Do an inversion with respect to the incircle. The image of $X$ is denoted as $X'$, etc. Let $X,Y,Z$ denote the tangency points of the incircle to $BC,CA,AB$. Let $QR$ touch the incircle at $W$. It is clear that $W,X,Y,Z$ are not moved and that $Q',R', A',B',C'$ are the midpoints of $Z'W',W'Y',Y'Z',Z'X',X'Y'$. Also note that $APYIZ$ is cyclic because those points lie on the circle with diameter $AI$. Consequently, $P'$ lies on the image of that circle which must be a line through $Y'Z'$. Also $I,P,P'$ are collinear. Further, it is obvious that $I,W,P$ are collinear, so $P'=Y'Z'\cap IW'$. Now observe that $\triangle Z'X'P'\sim\triangle W'Y'P'$ and $B'$, $R'$ are midpoints of $Z'X'$ and $W'Y'$, so we can easily conclude that $\triangle B'X'P\sim\triangle R'Y'P'$, so $\angle P'B'X'=\angle Y'R'P'$. It is clear that $\angle IB'X'=\angle IR'Y'$, so subtracting from above, we get $\angle P'B'I=\angle P'R'I$. $\star$. By symmetry, we also have $\angle P'C'I=\angle P'Q'I$. Subtracting these relations gives us $\angle P'Q'I-\angle P'B'I=\angle P'C'I-\angle P'R'I$ In general for an inversion, $\angle RSI=\angle S'R'I$ where $I$ is the center of inversion and $R',S'$ denote the images of $R,S$, so using that above, we get $\angle QPI-\angle BPI=\angle CPI-\angle RPI$ which of course reduces to the desired result of $\angle QPB=\angle CPR$. As a remark, this solution is pretty straightforward application of inversion, you just need to observe in the inverted diagram how to prove the angle equality.

Since $\omega$ is an excircle to $AQR$, we have $QF=ER$ Then $\frac{QF}{BD}=\frac{AF}{FD}=\frac{PE}{ED}=\frac{ER}{SD}$ Thus $SD=BD$ or $\triangle PBS$ is isosceles . Now, $\frac{QJ}{AP}=\frac{AQ}{QB}=\frac{AR}{RC}=\frac{RH}{AP}$ Thus, $QJ=RH$ And $\frac{QJ}{GB}=\frac{PJ}{BJ}=\frac{PR}{RC}=\frac{RH}{SC}$ Thus, $\triangle PGC $ is isosceles and we are done.