jmc

Since $RPS$ is a straight line, then $\angle SPQ=180^{\circ }-\angle RPQ=180^{\circ }-3y^{\circ }$.

Using the angles in $△PQS$, we have $\angle PQS+\angle QSP+\angle SPQ=180^{\circ }$. Thus, $x^{\circ }+2y^{\circ }+(180^{\circ }-3y^{\circ })=180^{\circ }$ or $x-y+180=180$ or $x=y$.

(We could have instead looked at $\angle RPQ$ as being an external angle to $△SPQ$.)

Since $x=y$, then $\angle RQS=2y^{\circ }$.

Since $RP=PQ$, then $\angle PRQ=\angle PQR=x^{\circ }=y^{\circ }$. Therefore, the angles of $△RQS$ are $y^{\circ }$, $2y^{\circ }$ and $2y^{\circ }$.

Thus, $y^{\circ }+2y^{\circ }+2y^{\circ }=180^{\circ }$ or $5y=180$ or $y=36$.

Therefore, $\angle RPQ=3y^{\circ }=3(36{)}^{\circ }=108^{\circ }$, so our final answer is $\boxed{108}$ degrees.