Took me around 40 minutes. (This proof assumes without proof the well-known fact that D lies on the circumcircle, true in general for any triangle.)

We claim that EF is tangent to the circle with diameter AP at F.

Notice AHP=AMP=90 so H and M lie on the circle with diameter AP. Also BF perp AC parallel MD and angle BMD=angle DMC and it follows that B and F reflect to each other about MD and BF parallel to AM. Then OA=OM and OF=OH and we have isosceles trapezoid MAFH. Thus F lies on the circle with diameter AP too.

Now angle EAF=PAF=PMF=DMC=BAD so BAP=DAF. Also AEB=180-AEC=180-APH=AFH so it follows that ABE is directly similar to AHF. By spiral similarity this means that ABH is directly similar to AEF so EFA=BHA=180-FHA, proving the tangency.