Circles $\omega,\gamma$, centered at $O$, have radii $28,8\sqrt3$. Equilateral $\triangle ABC$ has $A$ on $\omega$, $\overline{BC}$ tangent to $\gamma$, interior in $\omega$ but outside $\gamma$. $\overline{AO},\overline{BC}$ meet at $P$ with $BP/CP=5/3$. What is $AB$?

1. With $AB=a$: $BD=\frac a2, BP=\frac58a, DP=\frac18a, AD=\frac{\sqrt3}2a$, and $PE:PD=OE:AD$ gives $PE=2$.
2. Using $AP+OP=28$ (the sum of segments along $\overline{AO}$ out to the radius): $\sqrt{\frac34a^2+\frac1{64}a^2}+\sqrt{(8\sqrt3)^2+4}=28$.
3. Solving gives $a=\boxed{16}$.