Sol:- Let $AP$ meet $(AEF)$ again at $J$. Feet of $D$ on $AB,AC$ be $E',F'$. $E'-M-F'$ is simson line of $D$ . $AD \cap E'F'=K$. Note that $AEJFP$ and $AE'DF'K$ are homothetic . So $\frac{AP}{AJ}=\frac{AK}{AD} \implies AP \cdot AD=AJ \cdot AK$. So inverting at $A$ with radius $\sqrt(AP \cdot AD)$ maps $J$ to $K$ and hence $(AEF)$ to the line through $K$ perpendicular to $AJ$ i.e. line $E'-K-M-F$. $(PMD)$ is mapped to itself. So $M=(PMD) \cap E'F' \cap AM$. If $M'$ denotes the image of $M$ under this inversion we get $(PMD) \cap (AEF) \cap AM =M'$. So done

Another solution : Let $K$ be the intersection of $(AEF)$ and $(ABC)$ Claim: $K$ is on $(PMD)$ The solution is quite simple: we observe the similar triangles $KEB \sim KFC$ so we have (since P is the midpoint of EF) $KEF \sim KBC$ so $KEB \sim KFC \sim KPM$ so $\angle KMP=\angle KCF=\angle KCA=\angle KDA=\angle KDP$ so $KDMP$ is cyclic Finish : Let J be the intersection of $AM$ and $(KPMD)$. We want J to be on $(AEF)$ We have that $\angle KJA=180-\angle KJM=180-\angle KPM=180-\angle KFC=\angle KFA$ so $J$ is on $(KAF)$ and we are done

More on this configuration, I will use the points in the diagram of the #2 post: $BJ,CJ$ cut $AC,AB$ at $U,V$. $(AUV)$ cut $(ABC)$ again at $S$. $N$ midpoint arc $BAC$ of $(ABC)$. $N'$ reflects $N$ over $BC$. Then $S,J,D,N'$ lie on a circle.

Let circle $(AEF)$ intersect $AD$ again at point $R$ and $AM$ at point $S$. Moreover, let the Simson line of point $M$ wrt circle $(ABC)$ intersect $AB,AC$ at points $X,Y$ respectively. Then, if $XY$ intersects $AD$ at point $T$, we have $\angle ATM=90^\circ$ and $\angle ASR=90^\circ, $ and so $RSMT$ is cyclic. Moreoever, $ERTX$ and $EPDX$ are cyclic too, since $\angle AER=\angle ATX=90^\circ$ and $\angle APE=\angle AXD=90^\circ$. Thus, $AS \cdot AM=AR \cdot AT=AE \cdot AX=AP \cdot AD,$ which implies that $PSMD$ is cyclic, as desired.

Suppose $V = \odot(AEF)\cap \odot(ABC),U=\odot(AEF)\cap\odot(PMD)$ Consider that $P$ is the mid-point of $EF$,and $\triangle VEF\sim \triangle VBC$ So $\triangle VEP\sim \triangle VBM$ Hence $\angle VPD = \angle VPE + 90^{\circ} = \angle VMD$ So $V,P,M,D$ are cyclic Consider $\angle VUM = 180^{\circ} - \angle VDM = \angle VAD - 90^{\circ} = 180^{\circ} - \angle AEV = 180^{\circ} - \angle AUV$ Which means that $A,U,M$ are collinear So $AM$ and the circumcircles of the triangles $AEF$ and $PMD$ pass through a common point

I don't know why this was put as the hardest geo on the shortlist. Anyway, here's a solution: Let $K$ and $L$ be the second intersections of the circumcircle of $\triangle AFE$ with $\overline{AM}$ and $\overline{AD}$ respectively. Let $S$ and $K'$ be the second intersections of the circumcircle of $\triangle ABC$ with $\overline{DM}$ and $\overline{AM}$ respectively. Clearly, $S$ is the midpoint of arc $BAC$ in that circumcircle, and direct angle chasing shows that $SCK'DB\sim AFKLE$, so $\angle AKP = \angle SK'M = \angle MDA$. Hence $MDPK$ is cyclic and we're done as the desired intersection point is $K$.

Solved with mueller.25 Let $K$ be the intersection of $(AEF)$ and $(ABC)$. Then $\angle KXM = \angle KBE = \angle KBA = \angle KDA = \angle KDP$, so $K$ lies on $(PMD)$. To finish, note that if $D'$ is the antipode of $A$ on $(AEF)$. Then $\angle KXM = 180 - \angle KDM = 180 - \angle KD'P = 180 - \angle KD'A = 180 - \angle KXA$, so points $A,X,M$ are collinear, so we're done. $\blacksquare$

In this solution, I only use angle chasing after adding new points namely $X,Y,Z,K,L.$

Great Problem! Let $E_1,F_1$ be the feet of perpendicular from $D$ to lines $AB$ and $AC$, respectively. From Fact 5, we have that $DM\perp BC$. Since $D\in(ABC)$, points $E_1,M,F_1$ are collinear (Simson Line). Moreover, $\angle DPE=90^{\circ}=\angle DE_1A=\angle DE_1E\Longrightarrow E_1,D,P$ are cyclic. So, $$AE_1\cdot AE=AP\cdot AD=AF_1\cdot AF$$Consider an inversion with center $A$ and radius $\sqrt{AP\cdot AD}$. Under this inversion, $E\longleftrightarrow E_1$, $F\longleftrightarrow F_1$, $P\longleftrightarrow Q$. Let $K\ne M$ be the intersection of $AM$ with $(PMD)$. Then, $AP\cdot AD=AM\cdot AK$ so $K\longleftrightarrow M$ under inversion. Also, line $E_1F_1\longleftrightarrow(AEF)$. So, $K\in(AEF)$, as desired. $\square$

Imagine this actly being a G6 bruh. We first have the following key claim. Claim : Circles $(ABC),(AEF)$ and $(PMD)$ pass through a common point. Proof : Let $S=(AEF)\cap (ABC)$. Then, $S$ is the center of the spiral similarity $EF \mapsto BC$. Let $M_A$ be the minor arc midpoint of $\widehat{EF}$ (not containing $A$). Then, clearly $A-P-M_A-D$ (all these point must lie on the angle bisector of $\angle BAC$). Further, this spiral similarity maps $P$ to $M$ (midpoints) and $M_A$ to $D$ (minor arc midpoints). Thus, we must have $\triangle SPM_A \sim \triangle SMD$. From which we have, \[\measuredangle DPS = \measuredangle M_APS = \measuredangle DMS\]Thus, $S$ also lies on $(PMD)$. Now, let $G = (AEF) \cap (PMD) \neq S$. Then, one sees that \begin{align*} \measuredangle SGA &= \measuredangle SFA\\ &= \measuredangle SAF + \measuredangle FEA\\ &= \measuredangle SAC + \measuredangle FEA\\ &= \measuredangle CBS + \measuredangle FEA\\ &= \measuredangle CSB + \measuredangle BCS + \measuredangle AEF\\ &= \measuredangle CAB + \measuredangle BCA + \measuredangle ACS + \measuredangle AEF \end{align*}Further, let $L$ be the midpoint of the arc $\widehat{BAC}$. \begin{align*} \measuredangle MGS &= \measuredangle SDM\\ &= \measuredangle SDL\\ &= \measuredangle SCL\\ &= \measuredangle SCA + \measuredangle ACL\\ &= \measuredangle SCA + \measuredangle BCL + \measuredangle ACB \end{align*}Now, one can note that clearly $\measuredangle AEF = \measuredangle BCL$ (angles of isosceles triangles with vertex $\angle A$). Thus, \begin{align*} \measuredangle SGA + \measuredangle MGS &= \measuredangle CAB + \measuredangle BCA + \measuredangle ACS + \measuredangle AEF + \measuredangle SCA + \measuredangle BCL + \measuredangle ACB\\ &= \measuredangle CAB + ( \measuredangle AEF + \measuredangle BCL)\\ &= \measuredangle CAB + \measuredangle BAC\\ & = 0 \end{align*}Thus, the only possibility is that $A-G-M$ and thus, $\overline{AM}$ passes through one of the intersection points of $(AEF)$ and $(PMD)$ as claimed.

Solution In the figure below, define $G$ to be the intersection of circle $(AEF)$ and line $(AM)$. Define $L$ as the second intersection of line $(BC)$ and circle $(PMD)$. Also, define $H$ as the second intersection of circle $(ABC)$ and circle $(PMD)$. We clearly see that $L$ is the antipode of $D$ in $(PMD)$ and $P,E,F,L$ are colinear since $DPL = 90$°. Claim : $HEBL$ is cyclic. Proof: $\angle HBE = \angle HBA = \angle HDA = \angle HDP = \angle HLP = \angle HLE$. Claim 2: $HEFA$ is cyclic. Proof : $\angle HEF = 180 - \angle HEL = 180 - \angle HBL = \angle HBC = 180 - \angle HAC = 180 - \angle HAF$. To conclude, notice that $$180 - \angle HGM = \angle HGA = \angle HEA = \angle BHE + \angle HBE = \angle BLE + \angle HLE = \angle BLH = \angle MLH = \angle MDH$$ Hence $G$ lies on $(HMD) = (PMD)$, as desired.

Let $(ABC)$ intersect line $AM$ at $L$, and intersect $(AEF)$ at $S$. Let $X$ be the intersection of $(AEF)$ with $AM$. Let $N$ be the midpoint of arc $BAC$. Note that $D,M,N$ are collinear. We need to show that $(PXMD)$ is concyclic. Consider the spiral similarity centred at $S$ taking $EF$ to $BC$. Note that this also takes $A$ to $N$, $P$ to $M$, and $X$ to $L$. Now $\measuredangle MXP = \measuredangle AXP = \measuredangle NLM = \measuredangle NLA = \measuredangle NDA = \measuredangle MDP$, so $(PXMD)$ is concyclic and we are done.