It's $30^\circ$.

Let $BM\cap AN=H$. Apply Pascal's theorem in hexagon $AN(N)BM(M)$, we know $C,P,H$ are collinear. And $AN\perp BC,BM\perp AC\Rightarrow H$ is the orthocenter of $ABC$. Then $CH\perp AB$. And $\angle PHN=\angle ABC=\angle PNH, \angle PHM=\angle PMH\Rightarrow O$ is the circumcircle center of $CMHN$. Thus, $CP=MP=NP=MN\Rightarrow \angle MPN=60^\circ \Rightarrow \angle MCN=30^\circ$. [asy][asy] pair a,b,c; a=(0,0); b=(5,0); c=(2,5); draw(a--b--c--cycle); pair o,m,n,p; o=a/2+b/2; m=foot(b,a,c); n=foot(a,b,c); p=extension(m,rotate(90,m)*o,n,rotate(90,n)*o); draw(circle(o,2.5)^^m--n--p--cycle^^c--p); pair h;h=extension(a,n,b,m); draw(a--n^^b--m^^p--h,dotted); dot(o); label("$A$",a,W); label("$B$",b,E); label("$C$",c,N); label("$M$",m,WNW); label("$N$",n,NE); label("$P$",p,NE); label("$H$",h,S); label("$O$",o,S); [/asy][/asy]

jasperE3 wrote: Let $ABC$ be an acute triangle. Circle $k$ with diameter $AB$ intersects $AC$ and $BC$ again at $M$ and $N$ respectively. The tangents to $k$ at $M$ and $N$ meet at point $P$. Given that $CP=MN$, determine $\angle ACB$. Here is a slightly different solution Let $\omega$ and $\gamma$ denote the circles circumscribing $CMN$ and $ABMN$,also denote $D$ as the centre of $\gamma$.Then note that $MN$ is the radical axis. So, line joining their centres must be perpendicular to $MN$. We now show that $P$ is the centre of $\omega$. Note that $DMPN$ is a kite. Hence, diagonals $DP$ and $MN$ are perpendicular. Hence, $P$ is the centre of $\omega$. So, $CP=MP=PN=MN$, hence $\triangle MNP$ is equilateral $\implies MPN=60^{\circ}$.Hence by inscribed angle theorem, $\angle ACB = 30^{\circ} \blacksquare$