smc

Because $PA$ is a tangent line, angle $\angle OAP$ is a right angle. Drop a perpendicular from $O$ to $BC$ at $E.$ We find that $BE=EC=10.$ Let $AR=r$ and $PE=a$. We now have a system of equations. $$(10\sqrt{3}{)}^2+r^2=P{R}^2$$ $$(\sqrt{r^2-10^2}{)}^2+P{E}^2=r^2-10^2+P{E}^2=P{R}^2$$ Set them equal to each other and solve. $$(10\sqrt{3}{)}^2+r^2=r^2-10^2+P{E}^2$$ $$(10\sqrt{3}{)}^2=-10^2+P{E}^2$$ $$400=P{E}^2$$ $$20=PE$$ We know $BE=10$, so $PE=20-10=10$, which is $\boxed{10}$.