jmc

Since $PA+PD=8,$ point $P$ must lie on the ellipse whose foci are $A$ and $D,$ and whose major axis has length $8.$ Since the distance between the foci is $3-(-3)=6,$ the minor axis has length $\sqrt{8^2-6^2}=2\sqrt{7}.$ Then the semi-axes have lengths $4$ and $\sqrt{7},$ respectively, and the center of the ellipse is $(0,0),$ so the equation of this ellipse is $$\frac{x^2}{16}+\frac{y^2}{7}=1.$$Similarly, since $PB+PC=8,$ point $P$ must lie on the ellipse whose foci are $B$ and $C,$ and whose major axis has length $8.$ Since the distance between the foci is $2-(-2)=4,$ the minor axis has length $\sqrt{8^2-4^2}=4\sqrt{3}.$ Then the semi-axes have lengths $4$ and $2\sqrt{3},$ respectively, and the center of the ellipse is $(0,1),$ so the equation of this ellipse is $$\frac{x^2}{16}+\frac{(y-1{)}^2}{12}=1.$$Both ellipses are shown below. (Note that they intersect at two different points, but that they appear to have the same $y-$coordinate.) Since $P$ lies on both ellipses, it must satisfy both equations, where $P=(x,y).$ We solve for $y.$ By comparing the two equations, we get $$\frac{y^2}{7}=\frac{(y-1{)}^2}{12}.$$Cross-multiplying and rearranging, we get the quadratic $$5y^2+14y-7=0,$$and so by the quadratic formula, $$y=\frac{-14\pm \sqrt{14^2+4\cdot 5\cdot 7}}{10}=\frac{-7\pm 2\sqrt{21}}{5}.$$It remains to determine which value of $y$ is valid. Since $\sqrt{21}>4,$ we have $$\frac{-7-2\sqrt{21}}{5}<\frac{-7-2\cdot 4}{5}=-3.$$But the smallest possible value of $y$ for a point on the ellipse $\frac{x^2}{16}+\frac{y^2}{7}=1$ is $-\sqrt{7},$ which is greater than $-3.$ Therefore, we must choose the $+$ sign, and so $$y=\frac{-7+2\sqrt{21}}{5}.$$The final answer is $7+2+21+5=\boxed{35}.$