jmc

The circle with center $C$ is the incircle of $△PQR$. So, any segment from a vertex of the triangle to $C$ is an angle bisector.

From $\angle QRC=30^{\circ }$, we find that $\angle QRP=60^{\circ }$ because $RC$ is an angle bisector.

The sum of the measures of the internal angles of a triangle is $180^{\circ }$, so

$$\begin{array}{rl}\angle QPR& =180^{\circ }-\angle PQR-\angle QRP\\ & =180^{\circ }-65^{\circ }-60^{\circ }.\end{array}$$This yields $\angle QPR=\boxed{55^{\circ }}$.