jmc

Both triangles $APD$ and $CBD$ are 30-60-90 triangles. Thus $DP=\frac{2\sqrt{3}}{3}$ and $DB=2$. Since $\angle BDP=\angle PBD$, it follows that $PB=PD=\frac{2\sqrt{3}}{3}$. Hence the perimeter of $△BDP$ is $\frac{2\sqrt{3}}{3}+\frac{2\sqrt{3}}{3}+2=2+\frac{4\sqrt{3}}{3}$. Converting this to our contrived notation gives $w+x+y+z=\boxed{12}$.