jmc

Let $O$ be the circumcenter of $ABC$ and let the intersection of $CP$ with the circumcircle be $D$. It now follows that $\angle DOA=2\angle ACP=\angle APC=\angle DPB$. Hence $ODP$ is isosceles and $OD=DP=2$. Denote $E$ the projection of $O$ onto $CD$. Now $CD=CP+DP=3$. By the Pythagorean Theorem, $OE=\sqrt{2^2-\frac{3^2}{2^2}}=\sqrt{\frac{7}{4}}$. Now note that $EP=\frac{1}{2}$. By the Pythagorean Theorem, $OP=\sqrt{\frac{7}{4}+\frac{1^2}{2^2}}=\sqrt{2}$. Hence it now follows that, $$\frac{AP}{BP}=\frac{AO+OP}{BO-OP}=\frac{2+\sqrt{2}}{2-\sqrt{2}}=3+2\sqrt{2}$$ This gives that the answer is $\boxed{7}$.