smc

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 $\boxed{5}$.