Let the circumcircle of triangle $O B P$ intersect side $B C$ at the points $R$ and $B$ and let $\angle A, \angle B$ and $\angle C$ denote the angles at vertices $A, B$ and $C$, respectively.

Now note that since $\angle B O P=\angle B$ and $\angle C O Q=\angle C$, it follows that

$\angle P O Q=360^{\circ}-\angle B O P-\angle C O Q-\angle B O C=360^{\circ}-(180-\angle A)-2 \angle A=180^{\circ}-\angle A$.

This implies that $A P O Q$ is a cyclic quadrilateral. Since $B P O R$ is cyclic,

$\angle Q O R=360^{\circ}-\angle P O Q-\angle P O R=360^{\circ}-\left(180^{\circ}-\angle A\right)-\left(180^{\circ}-\angle B\right)=180^{\circ}-\angle C$.

This implies that $C Q O R$ is a cyclic quadrilateral. Since $A P O Q$ and $B P O R$ are cyclic,

$\angle Q P R=\angle Q P O+\angle O P R=\angle O A Q+\angle O B R=\left(90^{\circ}-\angle B\right)+\left(90^{\circ}-\angle A\right)=\angle C$.

Since $C Q O R$ is cyclic, $\angle Q R C=\angle C O Q=\angle C=\angle Q P R$ which implies that the circumcircle of triangle $P Q R$ is tangent to $B C$. Further, since $\angle P R B=\angle B O P=$ $\angle B$,

$$\angle P R Q=180^{\circ}-\angle P R B-\angle Q R C=180^{\circ}-\angle B-\angle C=\angle A=\angle P A Q$$

This implies that the circumcircle of $P Q R$ is the reflection of $\Gamma$ in line $P Q$. By symmetry in line $P Q$, this implies that the reflection of $B C$ in line $P Q$ is tangent to $\Gamma$.