Solution:
Let the center of $\Gamma$ be $O$, the radius $r$. Since $BP = BC$, let $\theta = \measuredangle BPC = \measuredangle BCP$.

Quadrilateral $QABC$ is cyclic, so $\measuredangle BAQ = 180^\circ - \theta$ and hence $\measuredangle PAQ = 120^\circ - \theta$.

Also $\measuredangle APQ = 180^\circ - \measuredangle APB - \measuredangle BPC = 120^\circ - \theta$, so $PQ = AQ$ and $\measuredangle AQP = 2\theta - 60^\circ$.

Again because quadrilateral $QABC$ is cyclic, $\measuredangle ABC = 180^\circ - \measuredangle AQC = 240^\circ - 2\theta$.

Triangles $OAB$ and $OCB$ are congruent, since $OA = OB = OC = r$ and $AB = BC$.

Thus $\measuredangle ABO = \measuredangle CBO = \frac{1}{2} \measuredangle ABC = 120^\circ - \theta$.

We have now shown that in triangles $AQP$ and $AOB$, $\measuredangle PAQ = \measuredangle BAO = \measuredangle APQ = \measuredangle ABO$.

Also $AP = AB$, so $\triangle AQP \cong \triangle AOB$. Hence $QP = OB = r$.

Let the center of $\Gamma$ be $O$, the radius $r$. Since $A, P$ and $C$ lie on a circle centered at $B$, $60^\circ = \measuredangle ABP = 2 \measuredangle ACP$, so $\measuredangle ACP = \measuredangle ACQ = 30^\circ$.

Since $Q, A$, and $C$ lie on $\Gamma$, $\measuredangle QOA = 2 \measuredangle QCA = 60^\circ$.

So $QA = r$ since if a chord of a circle subtends an angle of $60^\circ$ at the center, its length is the radius of the circle.

Now $BP = BC$, so $\measuredangle BPC = \measuredangle BCP = \measuredangle ACB + 30^\circ$.

Thus $\measuredangle APQ = 180^\circ - \measuredangle APB - \measuredangle BPC = 90^\circ - \measuredangle ACB$.

Since $Q, A, B$ and $C$ lie on $\Gamma$ and $AB = BC$, $\measuredangle AQP = \measuredangle AQC = \measuredangle AQB + \measuredangle BQC = 2 \measuredangle ACB$.

Finally, $\measuredangle QAP = 180 - \measuredangle AQP - \measuredangle APQ = 90 - \measuredangle ACB$.

So $\measuredangle PAQ = \measuredangle APQ$ hence $PQ = AQ = r$.