smc

We claim that $\boxed{A}$ is the right answer. Let the radius of circle $O$ be $r$, and let the length of $AD=x$. Since $AB\perp BC$, $AB$ is a tangent to circle $O$. Thus, by the tangent-secant theorem, we have $A{B}^2=AD\times AE$, or, $(2r{)}^2=x(x+2r)$. Through some algebraic manipulation, we find $$\begin{array}{r}A{D}^2=x^2=2r(2r-x)\end{array}$$. Since $A{D}^2=A{P}^2=x^2$, $PB=AB-AP=2r-x$, and $AB=2r$, we see that $(1)$ is identical to $A{P}^2=PB\times AB$, hence our answer is $\boxed{A{P}^2=PB\times AB}$