Very easy. We have to prove that $\angle ABC =\angle APM$ $H$ is the orthocenter of triangle $ABC$, so $\angle DHB=\angle ABC$ But $\angle ABC =\angle DHB=\angle XHA=\angle XPA=\angle PAB$. Then $PDBA$ is an isosceles trapezoid, hence $\angle APM=\angle BDM=\angle BCA=\angle ABC$, as claimed.

Since $DP\parallel BA$ and $D$ is the midpoint of $BC$, then $DP$ intersects $AC$ at the midpoint of $AC$, say $N$. We have $\angle APN=\angle AHX=\angle ACB=\angle AMN$, so $AMPN$ is cyclic. The circumcenter of $AMP$, say $O$, clearly lies on $AH$. Thus the tangent of $(AMP)$ at $A$ is perpendicular to $AO$, which is parallel to $AH$, which is perpendicular to $BC$.

Let $\angle{ABC}=\angle{ACB}=\theta$.Then $\angle{BHD}=\theta=\angle{DPA}=\angle{DBA} \implies ABPD$ is cyclic.Also see that ${{\angle{BDP}=\angle{NDC}=\angle}NCD}=\theta=\angle{DBA}$ so $ABPD$ is an isosceles trapezoid.Now some trivial angle chasing gives you $\angle{PBA}=90-\theta$ so $\angle{AMP}=180-2\theta=\angle{PAX}$ where $AX$ is the tangent line from $A$ to the circumcircle of $AMP$.Thus $\angle{DAX}=\angle{PAX}-\angle{PAD}=180-2\theta-90+2\theta=90^{\circ}=\angle{ADB} \implies AX \parallel BC$.

Let $N$ be the mid point of $AC$. Note that it suffices to proving that if $\angle ABC=\angle ACB=\theta$ then $\angle APN=\theta$. Now this is trivial since $\angle APN=\angle APX=\angle AHX=\theta$. Lol. Too trivial for Iran.

Such a disappointment We want to show, $\odot (AMP)$ and $\odot (ABC)$ are internally tangent which is equivalent to show, $\odot (APM)$ bisects $AC$. Let $E$ be midpoint of $AC$ $$\angle APX=\angle AHX=\angle ACB=\angle ABC=\angle AME \qquad \blacksquare$$

[asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(10cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -6.38667656855066, xmax = 9.682272354136112, ymin = -1.3623798338963669, ymax = 8.424293845080454; /* image dimensions */ pen ttzzqq = rgb(0.2,0.6,0); pen ccqqqq = rgb(0.8,0,0); pen qqqqcc = rgb(0,0,0.8); pen yqqqyq = rgb(0.5019607843137255,0,0.5019607843137255); pen qqzzff = rgb(0,0.6,1); /* draw figures */ draw((0,8)--(-5,0), linewidth(2.4) + ttzzqq); draw((5,0)--(0,8), linewidth(2.4) + ttzzqq); draw(circle((0.7051282051282051,5.5625), 2.5374420260702166), linewidth(2) + ccqqqq); draw(circle((0,5.21875), 2.78125), linewidth(2) + qqqqcc); draw((0,8)--(2.1910112359550564,3.50561797752809), linewidth(2) + yqqqyq); draw((-2.5,4)--(2.1910112359550564,3.50561797752809), linewidth(1.6) + qqzzff); draw((-5,0)--(3.2051282051282053,5.128205128205129), linewidth(2.4) + red); draw((3.2051282051282053,5.128205128205129)--(0,0), linewidth(2) + red); draw((-5,0)--(0,0), linewidth(2.4) + ttzzqq); draw((0,0)--(5,0), linewidth(2.4) + ttzzqq); draw((0,8)--(0,0), linewidth(2.4) + red); /* dots and labels */ dot((0,8),dotstyle); label("$A$", (0.0522994181997128,8.139383403188845), NE * labelscalefactor); dot((5,0),dotstyle); label("$C$", (5.052477673397458,0.14764550812916896), NE * labelscalefactor); dot((0,3.125),linewidth(4pt) + dotstyle); label("$H$", (0.023808374010551855,2.8400491840048883), NE * labelscalefactor); dot((0,0),linewidth(4pt) + dotstyle); label("$D$", (0.13777255076719563,-0.2939656768028273), NE * labelscalefactor); dot((-2.5,4),linewidth(4pt) + dotstyle); label("$M$", (-2.896523655378445,3.8942178190038472), NE * labelscalefactor); dot((3.2051282051282053,5.128205128205129),linewidth(4pt) + dotstyle); label("$X$", (3.3572605441423824,4.877158843529903), NE * labelscalefactor); dot((2.1910112359550564,3.50561797752809),linewidth(4pt) + dotstyle); label("$P$", (2.3030919091434274,3.3386424573152067), NE * labelscalefactor); dot((-5,0),linewidth(4pt) + dotstyle); label("$B$", (-5.076088535849257,0.07641789765626635), NE * labelscalefactor); dot((2.191011235955056,4.49438202247191),linewidth(4pt) + dotstyle); label("$E$", (2.174882210292203,4.6777215342057765), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] To prove: Tangent from $A$ to the circumcircle of $\Delta AMP$ is parallel to $BC$. In other words, we need to show that $\angle ABC=\angle APM$. Since, $H$ is the orthocentre of the triangle $ABC$, we have $\angle DHB= \angle ABC$. Now, we can do Angle chasing for the angles made on the two circles. $\angle DHB=\angle ABC=\angle XHA=\angle XPA$. Now, by Angle Segment Formula, $\angle XPA=\angle PAM=\angle PAB$. Hence, we get $PDBA$ is an isosceles trapezoid. Using the condition $XD || AB$ we get, $\angle APM=\angle BDM=\angle BCA=\angle ABC$. Therefore, we get the final condition $\angle APM=\angle ABC. \qquad \blacksquare$

What kind of a tst problem is this ? Let $N$ be the mid point of $AC$.Then $\angle APN =\angle AHX = \angle C= \angle B =\angle AMN \implies (AMPN)$. Thus the angle the said tangent makes with $AB = \angle APM = \angle ANM = \angle C= \angle B$