The product of two lengths with a common point brings to mind the Power of a Point Theorem.
Since $PA = PB$, we can make a circle with radius $PA$ that is centered on $P$, and both $A$ and $B$ will be on that circle. Since $\angle APB = \widehat {AB} = 2 \angle ACB$, we can see that point $C$ will also lie on the circle, since the measure of arc $\widehat {AB}$ is twice the measure of inscribed angle $\angle ACB$, which is true for all inscribed angles.

Since $PDB$ is a line, we have $PD + DB = PB$, which gives $3 = DB + 2$, or $DB = 1$.
We now extend radius $PB = 3$ to diameter $EB = 6$. Since $EDB$ is a line, we have $ED + DB = EB$, which gives $ED + 1 = 6$, or $ED = 5$.
Finally, we apply the power of a point theorem to point $D$. This states that $AD \cdot DC = DB \cdot DE$. Since $DB = 1$ and $DE = 5$, the desired product is $5$, which is $\boxed{A}$.