jmc

Rotate $△ABC$ $90^{\circ }$ counterclockwise about $C$, and let $B^{\prime }$ and $P^{\prime }$ be the images of $B$ and $P$, respectively.



Then $C{P}^{\prime }=CP=6$, and $\angle PC{P}^{\prime }=90^{\circ }$, so $△PC{P}^{\prime }$ is an isosceles right triangle. Thus $P{P}^{\prime }=6\sqrt{2}$, and $B{P}^{\prime }=AP=11$. Because ${\left(6\sqrt{2}\right)}^2+7^2=11^2$, the converse of the Pythagorean Theorem implies that $\angle BP{P}^{\prime }=90^{\circ }$. Hence $\angle BPC=135^{\circ }$. Applying the Law of Cosines in $△BPC$ gives $$B{C}^2=6^2+7^2-2\cdot 6\cdot 7\cos 135^{\circ }=85+42\sqrt{2},$$and $a+b=\boxed{127}$.