Concurrence property of certain straight lines generated by a cyclic quadrilateral Cyclic quadrilateral $ABCD$ is inscribed in circle $\omega$. The opposite sides $AB$ and $CD$ intersect at $M$. Point $N$ is on $\omega$ such that $MN$ is tangent to the circle. $NB$ intersects $AC$ at point $P$. Given that $MN\parallel AC$, prove that $AN$, $DB$ and $PM$ concur.
My ideas so far: Let $R$ be the intersection of the circumcircle of $\triangle$$ABP$. $R$ is the radical center but I am not sure what the three circles are.

Comment. First, how difficult is it to make sure that $MN \, || \, AC$? Simply pick a point $M$ outside the circle $\omega$, draw one of the two tangents to $\omega$ through $M$, and denote by $N$ the point tangency of that line with $\omega$. Then, pick an arbitrary point $A$ on $\omega$, draw the line through $A$ parallel to $MN$, and let it intersect the circle $\omega$ for the second time at $C$. Thus, $B$ is the second intersection point of $AM$ with $\omega$ and $D$ is the second intersection point of $CM$ with $\omega$. Finally, $P$ is the intersection point of $NB$ with $AC$. 
Proof. Let $K$ be the intersection point of $MN$ and $BD$. Then $$\angle \, KDM = \angle \, BDC = \angle \, BAC = \alpha$$ as angles inscribed in $\omega$. On the other hand, since $MN \, ||\, AC$
$$\angle \, KMB = \angle \, BAC = \alpha$$ Hence $\angle \, KMB = \alpha = \angle \, KDM$. Combined with the fact that $\angle \, MKB = \angle \, DKM$, the two triangles $KBM$ and $KMB$ are similar. Consequently
$$\frac{KM}{KD} = \frac{KB}{KM}$$  which yields $$KM^2 = KB \cdot KD$$ However, due to the fact that $KN$ is a tangent of $\omega$ and $KD$ is a secant, 
$$KN^2 = KB \cdot KD$$ which means that $KM^2 =  KB \cdot KD = KN^2$. The latter identity is possible only when $KM = KN$, i.e. $K$ is the midpoint of $MN$. 
Finally, observe that $APMN$ is a trapezoid. Let $Q$ be the intersection point of $AN$ and $PM$. Then, in any trapezoid, the line connecting the point $Q$ to the intersection point $B$ of the diagonals $AM$ and $PN$ of $APMN$ must necessarily pass though the midpoint $K$ of the side $MN$ (as well as through the midpoint of $AP$). Therefore, the point $K$ lies on $QB$, meaning that the line $KB$ passes through $Q$. However, by construction, point $D$ also lies on $KB$ which means that the line $DB$, being the same as line $KB$, passes through the intersection point $Q$ of $AN$ and $PM$.    
