APMO

From the assumption that the circle $\Gamma$ intersects both of the line segments $AB$ and $AC$, it follows that the 4 points $N,C,Q,O$ are located on $\Gamma$ in the order of $N,C,Q,O$ or in the order of $N,C,O,Q$. The following argument for the proof of the assertion of the problem is valid in either case. Since $\angle NQC$ and $\angle NOC$ are subtended by the same arc $\text{overparen}NC$ of $\Gamma$ at the points $Q$ and $O$, respectively, on $\Gamma$, we have $\angle NQC=\angle NOC$.

We also have $\angle BOC=2\angle BAC$, since $\angle BOC$ and $\angle BAC$ are subtended by the same arc $\text{overparen}BC$ of the circumcircle of the triangle $ABC$ at the center $O$ of the circle and at the point $A$ on the circle, respectively. From $OB=OC$ and the fact that $ON$ is a diameter of $\Gamma$, it follows that the triangles $OBN$ and $OCN$ are congruent, and therefore we obtain $2\angle NOC=\angle BOC$. Consequently, we have $\angle NQC=\frac{1}{2}\angle BOC=\angle BAC$, which shows that the 2 lines $AP,QN$ are parallel.

In the same manner, we can show that the 2 lines $AQ,PN$ are also parallel. Thus, the quadrilateral $APNQ$ is a parallelogram.