jmc

Let $AP=x$ so that $PB=92-x.$ Extend $AD,BC$ to meet at $X,$ and note that $XP$ bisects $\angle AXB;$ let it meet $CD$ at $E.$ Using the angle bisector theorem, we let $XB=y(92-x),XA=xy$ for some $y.$ Then $XD=xy-70,XC=y(92-x)-50,$ thus$$\frac{xy-70}{y(92-x)-50}=\frac{XD}{XC}=\frac{ED}{EC}=\frac{AP}{PB}=\frac{x}{92-x},$$which we can rearrange, expand and cancel to get $120x=70\cdot 92,$ hence $AP=x=\frac{161}{3}$. This gives us a final answer of $161+3=\boxed{164}$.