smc

Consider the reflection $P^{\prime }$ of $P$ over the perpendicular bisector of $\overline{BC}$, creating two new isosceles trapezoids $DAP{P}^{\prime }$ and $CBP{P}^{\prime }$. Under this reflection, $P^{\prime }A=PD=4$, $P^{\prime }D=PA=1$, $P^{\prime }C=PB=2$, and $P^{\prime }B=PC=3$. Since $DAP{P}^{\prime }$ and $CBP{P}^{\prime }$ are isosceles trapezoids, they are cyclic. Using Ptolemy's theorem on $DAP{P}^{\prime }$, we get that $(P{P}^{\prime })(AD)+(PA)(P^{\prime }D)=(A{P}^{\prime })(PD)$, so $$P{P}^{\prime }\cdot AD+1\cdot 1=4\cdot 4.$$ Then, using Ptolemy's theorem again on $CBP{P}^{\prime }$, we get that $(BC)(P{P}^{\prime })+(BP)(C{P}^{\prime })=(B{P}^{\prime })(CP)$, so $$P{P}^{\prime }\cdot BC+2\cdot 2=3\cdot 3.$$ Thus, $P{P}^{\prime }\cdot AD=15$ and $P{P}^{\prime }\cdot BC=5$; dividing these two equations and taking the reciprocal yields $\frac{BC}{AD}=\boxed{\frac{1}{3}}$.