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Let $P'=AA\cap BB$, let $K'=CP'\cap (ABC)$. By the symmedian lemma, we have that $CK'$ is the symmedian of triangle $BK'A$. We have\begin{align*} \measuredangle MOC&= 2\measuredangle BAC+\measuredangle ACB\\ &= \measuredangle BK'C+\measuredangle MK'A+ \measuredangle AK'B\\ &= \measuredangle MK'C\implies CMOK' \text{ is cyclic.} \end{align*}Therefore, $K\equiv K'$ and $P\equiv P'$. Hence, $\measuredangle PAK=\measuredangle ACK=\measuredangle MCB$. We are done.

[asy][asy] import graph; size(10cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.4) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -11.577851239669423, xmax = 2.1676033057851223, ymin = -5.59140495867768, ymax = 7.135867768595032; /* image dimensions */ pen wwqqcc = rgb(0.4,0,0.8); pen qqwuqq = rgb(0,0.39215686274509803,0); pen ffqqtt = rgb(1,0,0.2); draw((-6.8058740687648855,-0.33327984247567694)--(-6.472594226289209,-0.13915391124056262)--(-6.666720157524323,0.19412593123511432)--(-7,0)--cycle, linewidth(1) + red); draw((-4,0.385694607919935)--(-4.385694607919935,0.385694607919935)--(-4.385694607919935,0)--(-4,0)--cycle, linewidth(1) + red); /* draw figures */ draw((-5.92,4.64)--(-7,0), linewidth(1) + blue); draw((-7,0)--(-1,0), linewidth(1) + blue); draw((-1,0)--(-5.92,4.64), linewidth(1) + blue); draw(circle((-4,1.7474137931034484), 3.4718086013385276), linewidth(1) + wwqqcc); draw(circle((-8.455208333333331,0.8737068965517242), 4.540071038484399), linewidth(1) + linetype("2 2")); draw((xmin, -5.099202433810226*xmin-25.547278408156536)--(xmax, -5.099202433810226*xmax-25.547278408156536), linewidth(1) + qqwuqq); /* line */ draw((-4,ymin)--(-4,ymax), linewidth(1) + qqwuqq); /* line */ draw((-7,0)--(-4,-5.1504686729156335), linewidth(1) + red); draw((-4,-5.1504686729156335)--(-1,0), linewidth(1) + red); draw((-7,0)--(-4,1.7474137931034484), linewidth(1) + ffqqtt); draw((-4,1.7474137931034484)--(-1,0), linewidth(1) + red); /* dots and labels */ dot((-7,0),dotstyle); label("$A$", (-6.9233057851239685,0.19041322314049505), NE * labelscalefactor); dot((-1,0),dotstyle); label("$B$", (-0.9233057851239684,0.19041322314049505), NE * labelscalefactor); dot((-5.92,4.64),dotstyle); label("$C$", (-5.850578512396695,4.826776859504126), NE * labelscalefactor); dot((-4,0),linewidth(4pt) + dotstyle); label("$M$", (-3.923305785123968,0.15404958677685873), NE * labelscalefactor); dot((-4,1.7474137931034484),linewidth(4pt) + dotstyle); label("$O$", (-3.923305785123968,1.8995041322314021), NE * labelscalefactor); dot((-4.685279187817258,-1.6560913705583764),linewidth(4pt) + dotstyle); label("$K$", (-4.614214876033059,-1.518677685950412), NE * labelscalefactor); dot((-4,-5.1504686729156335),linewidth(4pt) + dotstyle); label("$P$", (-3.923305785123968,-5.009586776859499), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); [/asy][/asy] Consider an inversion around the circumcircle of $\triangle ABC$. Note that the points $A$, $B$, $C$ and $K$ remain fixed. Let $M'$ be the inverse of $M$. Since $COMK$ is cyclic, its inverse is the line $CK$, so $M'=CK\cap OM$. Then $M'=P$ and $\angle OMA = \angle OAM'= \angle OAP = 90º$. Similarly, $\angle OBP = 90º$. Therefore $P$ is the intersection of the tangents from $A$ and $B$ to the circumcircle of $\triangle ABC$. It's well-known that $CP$ is a symmedian of $\triangle ABC$. Hence $\angle PAK = \angle ACP = \angle MCB$, as desired.

Let $P'$ be the intersection of the tangents to $(ABC)$ at $A$ and $B$. Let $E$ be the midpoint of arc $AB$ not containing $C$. Also, let $N$ be the intersection of lines $AB$ and $CK$. We have $\angle BKN=\angle BKC=\angle BAC$, $\angle NKA=\angle CKA=\angle CBA$ and $\angle MKN=\angle MKC=-\angle COM=-\angle COE=-2\angle CBE=-2(\angle CBA +\angle ABE)=-2\angle CBA -(\angle ABE +\angle EAB)=-2\angle CBA +\angle BEA=-2\angle CBA - \angle ACB$. Hence, $\angle MKA=\angle MKN +\angle NKA=(-2\angle CBA -\angle ACB)+\angle CBA=-(\angle CBA +\angle ACB)=\angle BAC.$ This gives us that $K$ is the $K$-symmedian of $\Delta BKA$. Then, from the ratio lemma, $\frac{BN}{NA}=(\frac{BK}{KA})^2$. But from the law of sines we have $\frac{BK}{KA}=\frac{\sin\angle KAB}{\sin\angle ABK}=\frac{\sin\angle KCB}{\sin\angle ACK}=\frac{\sin\angle NCB}{\sin\angle ACN}$. Once again from the ratio lemma, $\frac{BN}{NA}=(\frac{BC}{AC})\frac{\sin\angle NCB}{\sin\angle ACN}\Rightarrow \frac{\sin\angle NCB}{\sin\angle ACN}=\frac{BN}{NA}(\frac{AC}{BC})$. Thus, $ \frac{BN}{NA}=(\frac{BN}{NA}\frac{AC}{BC})^2\Rightarrow \frac{BN}{NA}=\frac{BC}{AC}$. And by the converse of ratio lemma, we have that $CN$ is a $C$-symmedian of $\Delta ACB$. Therefore, $P$ lies on the intersection of the $C$-symmedian of $\Delta ACB$ and the perpendicular bisector of $\overline{AB}$, which coincides with $P'$, since $\Delta P'AB$ is isosceles(the fact that it is on the symmedian follow from some laws of sines). The angle chasing follows directly from the tangency and the isogonality.