jmc

Since $\angle QRP=120^{\circ }$ and $QRS$ is a straight line, then $\angle PRS=180^{\circ }-120^{\circ }=60^{\circ }$.

Since $\angle RPS=90^{\circ }$, then $△SRP$ is a $30^{\circ }$-$60^{\circ }$-$90^{\circ }$ triangle.

Therefore, $RS=2PR=2(12)=24$.

Drop a perpendicular from $P$ to $T$ on $RS$. Since $\angle PRT=60^{\circ }$ and $\angle PTR=90^{\circ }$, then $△PRT$ is also a $30^{\circ }$-$60^{\circ }$-$90^{\circ }$ triangle.

Therefore, $PT=\frac{\sqrt{3}}{2}PR=6\sqrt{3}$.

Consider $△QPS$. We may consider $QS$ as its base with height $PT$.

Thus, its area is $$\frac{1}{2}(6\sqrt{3})(8+24)=\boxed{96\sqrt{3}}.$$