To actually solve it, I later named the trapezoid's height as $h$, extended $AB$ to intersect the circle again at $Q$, named a certain length $d$, used the intersecting secants theorem with the legendary Pythagorean theorem that we all know, went through some heachache-inducing algebra crunching, and eventually got it with the help of a calculator.

But, I thought, there must be a way to solve this that isn't so egregiously unelegant! Right?

So I named some angles. $\alpha=\angle BAC=\angle ACD=\angle PBD$, $\beta=\angle ABP=\angle CAD$, $\gamma=\angle ACB=\angle PDB$, $\delta=\angle BDC=\angle BPC$, $\epsilon=\angle CBD=\angle CPD$, $\zeta=\angle MDP$.

But I couldn't find much similarity that would be of use. Only that $\triangle APB\sim\triangle BDP\sim\triangle ACD$, which leads to $\delta=\alpha+\beta$ and $AB=\frac25AC$.

Even more mysteriously, with GeoGebra I was able to find that $\angle MPB=\epsilon$! How is that, and would that be meaningful information? I even tried extending $BC$ to the point $F$ such that $\triangle ABC\sim\triangle CFP$ (i.e. $AFBP$ a cyclic quadrilateral), which would require $F,M,P$ to be collinear–which I haven't yet figured out how to prove.