Square $ABCD$ has sides of length $12$, and $M$ is the point on $\overline{CD}$ with $DM=4$. A circle with radius $4$ and center $M$ intersects a circle with radius $12$ and center $A$ at points $P$ and $D$. What is the distance from $P$ to $\overline{AD}$?

1. Connect $MP, MD, DP, AM$. The line of centers $AM$ is perpendicular to the common chord $DP$ and bisects it; let $AM$ meet $DP$ at $N$.
2. In the right triangle $ADM$ (right angle at $D$), $AD=12$ and $DM=4$, so $AM=\sqrt{12^2+4^2}=\sqrt{160}$. $DN$ is the altitude to the hypotenuse, so $AN=\dfrac{AD^2}{AM}=\dfrac{144}{\sqrt{160}}$ and $DN=\dfrac{AD\cdot DM}{AM}=\dfrac{48}{\sqrt{160}}$, hence $DP=2DN=\dfrac{96}{\sqrt{160}}$.
3. Let $x$ be the distance from $P$ to $AD$ (the dashed segment $PQ$, with $Q$ on $AD$). Then $[ADP]=\frac12\cdot DP\cdot AN=\frac12\cdot\dfrac{96}{\sqrt{160}}\cdot\dfrac{144}{\sqrt{160}}=\dfrac{216}{5}$.
4. Also $[ADP]=\frac12\cdot AD\cdot x=6x$, so $x=\boxed{\dfrac{36}5}$.