1997 AMC 12 Problem 26

Draw the circle centered at $P$ through $A$ and $B$

The central-angle condition puts $C$ on that circle, so apply intersecting chords at $D$

Because $PA=PB,$ draw their circle with center $P.$ The condition $\angle APB=2\angle ACB$ is the central-inscribed angle relation, so $C$ lies on the same circle. Along line $PB,$ the other circle intersection is $E$ with $PE=3.$ Power of $D$ gives $$\begin{aligned} AD\cdot CD&=DE\cdot DB\\ &=(3-2)(3+2)=5. \end{aligned}$$ Thus A is correct.