It is really easy. $|AP\parallel AD|=|AE|^{2}=|AM\parallel AI|$, where the first equality follows from the power of a point and the second from the similarity of $AEI$ and $AEM$. Conclusion follows.

AE and AF being tangents to the circle I and being concurrent on PD, it follows that $ DEPF$ is a harmonic quad, hence the tangents at D and P to circle I concur on EF, let's call $ T$ their common point. since $ M\in AI \cap EF$, we have $ MI\perp MT$, then $ DI\perp DT$ and $ PI\perp PT$, consequently $ D, M$ and $ P$ lie on the circle of diameter $ TI$, done. Note: This way follows that $ TI\perp AD$ (was this proof known for the issue?) Best regards, sunken rock

Consider The inversion center $ I$ power $ k$ ( $ k=ID^2$): $ {M}\rightarrow{A}, {P}\rightarrow{P}, {D}\rightarrow{D}$. Since we'll get the result

TomciO wrote: It is really easy. $|AP\parallel AD|=|AE|^{2}=|AM\parallel AI|$, where the first equality follows from the power of a point and the second from the similarity of $AEI$ and $AEM$. Conclusion follows. Well though according to u this is trivial but we need to show this before that A,M,I are collinear. Which is trivial by congruence of the triangles AFM,AEM. Then done.......by your proof

People is saying trivial when its an Ibero of 1990 and it was P4!. Show some respect to problem PLS! Anyways...taking polars w.r.t. incircle $\mathcal P_A=EF$ thus $M$ and $A$ are inverses w.r.t. incircle and now by inversion w.r.t. incirle we have $P \to P$, $M \to A$ and $D \to D$ thus the problem is equivalent to show that $A,P,D$ are colinear and we are done since that is a condition of the problem. @below BRUH

Trivial. By PoP, $$AK\cdot AI=AE^2=AD\cdot AP.$$

It equivalent to show that, $\angle DPI = \angle DMI $ Create $\angle DPK = \angle DJK$ on Circle as the image where $L,N = FM\cap PD, EM\cap JK$ claim #1 $\triangle ALN$ is isosceles. pf, By butterfly theorem $LM = MN$ and since $\angle FMA = \angle FMI = 90$ Imply that, $\triangle ALN$ is isosceles. claim #2 $PJ // LN$ pf, Suppose $O = PJ\cap AM$, $M^\prime = AM \cap DK$ Since $(M^\prime, O ; M, A)$ is harmonic bundle. \[(M^\prime, O ; M, A) = (P,J ; O, P_\infty) \;\; ;\text{Perspective with point } D\]Thus, O is midpoint of $PJ$. Apply with claim #1, $\angle AOJ = \angle AMN \Longleftrightarrow PJ // LN$ By claim #2, $\angle PJM = \angle LMD$ Since, $\angle DMI = 90-\angle LMD = 90-\angle PJD = 90-\angle PDB$ So, $90-\angle PDB = \angle IDP = \angle IPD$ Thus, $\angle PJM = \angle IPD$. $P,M,I,D$ Concyclic.

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Another overkill is to observe that $DP$ is a symmedian in $DEF$, so $EF$ is a symmedian in $DEP$, so $M$ is the magic point of the symmedian of $E$ in $DEP$, so $P,I,M,D$ are concyclic.