jmc

Since $PT$ and $RQ$ are parallel, then $2x^{\circ }=128^{\circ },$ so $x=64,$ so $\angle TPQ=64^{\circ }.$



Since $PT$ and $QR$ are parallel, then $\angle TPQ$ and $\angle PQR$ are supplementary. Thus, $\angle PQR+64^{\circ }=180^{\circ },$ so $\angle PQR=\boxed{116}\text{ degrees}.$