jmc

Draw line segment $DE$ such that line $DE$ is concurrent with line $BC$. Then, $ABED$ is an isosceles trapezoid so $AD=BE=10$, and $BC=8$ and $EC=2$. We are given that $DC=12$. Since $\angle CED=120^{\circ }$, using Law of Cosines on $△CED$ gives$$12^2=D{E}^2+4-2(2)(DE)(\cos 120^{\circ })$$which gives$$144-4=D{E}^2+2DE$$. Adding $1$ to both sides gives $141=(DE+1{)}^2$, so $DE=\sqrt{141}-1$. $△DAP$ and $△EBQ$ are both $30-60-90$, so $AP=5$ and $BQ=5$. $PQ=DE$, and therefore $AB=AP+PQ+BQ=5+\sqrt{141}-1+5=9+\sqrt{141}\to (p,q)=(9,141)\to \boxed{150}$.