There are different diagrams possible depending on relative positions of the circle, line and point $P$. In one case, $\angle PRV$ and $\angle PQV$ are equal and in the other are complementary.

$\angle PRV = 180^\circ - \angle PQV$ as $PQVR$ is a cyclic quadrilateral. $S$ is the intersection of $VR$ and the circle. $T$ is the intersection of the line $PQ$ and the circle.

$\angle PRV = \angle STP$. Thus $S$ is a fixed point on the circle and $S$ is on $VR$. Hence $VR$ always passes through $S$.

The other case is similar.

Let $X$ be the other point where $PQ$ meets the circle and let $W$ be the other point where the line through $X$ parallel to $QV$ meets the circle - clearly $W$ is a fixed point on the circle.

Now $PQVR$ is a cyclic quadrilateral. Therefore $\angle PRV = 180^\circ - \angle PQV$. Similarly $PXWR$ is a cyclic quadrilateral, so $\angle PRW = 180^\circ - \angle PXW$. But $\angle PXW = \angle PQV$. Therefore $\angle PRW = \angle PRV$. Hence $R$, $W$ and $V$ are collinear, as required. (Note that different configurations are possible depending on which side of the line $PQ$ the point $R$ lies. However, the arguments are essentially the same in all cases.)