Suppose circle $PMC$ and line $AN$ intersect at $S$. Also circle $BQN$ intersect with $AM$ at $T$. If we prove $M$, $N$ , $S$ and $T$ are on a circle then problem is done. You see $\triangle ACS$ and $\triangle TBA$ are similiar. \[ \frac{AS}{AT}=\frac{AC}{BT}=\frac{AC}{CN}\cdot\frac{BM}{BT}=\frac{\sin \angle ANM}{\sin \angle CAS}\cdot\frac{\sin \angle BTM}{\sin \angle AMN}=\frac{\sin \angle ANM}{\sin \angle AMN}=\frac{AM}{AN} \] So $AS\cdot AN=AM\cdot AT$. Therefore $N$, $M$, $S$, $T$ are on a circle.

I am trying to prove it with Area Ratios, but stucked... This problem is equal to $\frac{\triangle BQN}{\triangle CPM}=\frac{BQ}{CP}$. Anyone has any ideas?

It definitely cannot be solved by area ratio only, because an attempt via trigo shows that you have to use the formulae like $\sin(x+y)$, $\sin x+\sin y$, etc. This implies simply handling the ratios must not work. One has to deal with something like concyclicity, similarity, etc.

The area ratio of two triangles with $\angle{B}=\angle{B'}$ is $\frac{\triangle{ABC}}{\triangle{A'B'C'}}=\frac{BA\cdot BC}{B'A'\cdot B'C'}$. I am trying to do it in that way.

Let's go for a nonstandard proof: The reflection with respect to the perpendicular bisector of the segment BC maps the point B to the point C and vice versa; also, it maps the point M to the point N and vice versa (in fact, the points M and N lie on the segment BC and satisfy BM = CN; this means that they are symmetric to each other with respect to the midpoint of the segment BC, and thus with respect to the perpendicular bisector of the segment BC as well). Let D be the image of the point A under this reflection, and let R be the image of the point P under this reflection. Altogether, the reflection with respect to the perpendicular bisector of the segment BC maps the points B, C, M, N, A, P to the points C, B, N, M, D, R, respectively. Hence, as the point P lies on the segment AN, its image R lies on the segment DM, and we have < RNB = < PMC, < MDB = < NAC and < RBC = < PCB (since reflections are congruence transformations and keep angles invariant - non-directed angles at least). Since the points N and D are the reflections of the points M and A in the perpendicular bisector of the segment BC, the quadrilateral AMND is an isosceles trapezoid; thus, it has a circumcircle. Let the line NQ meet the line BD at a point U, and let the line NR meet the line AB at a point V. Since < UNM = < QNB = < NAC = < MDB = < UDM, the points U, M, N, D lie on one circle, i. e. the point U lies on the circumcircle of the quadrilateral AMND. Since < VNM = < RNB = < PMC = < MAB = < VAM, the points V, M, N, A lie on one circle, i. e. the point V also lies on the circumcircle of the quadrilateral AMND. Altogether, we see that both points U and V lie on the circumcircle of the quadrilateral AMND, so the hexagon AMDUNV is cyclic; thus, according to the Pascal theorem, the points $AM\cap UN$, $MD\cap NV$ and $DU\cap VA$ are collinear. Since $AM\cap UN=Q$, $MD\cap NV=R$ and $DU\cap VA=B$, this means that the points Q, R and B are collinear. Consequently, < QBC = < RBC. Since < RBC = < PCB, this yields < QBC = < PCB, and the problem is solved. Darij

Shobber i could not understand what do you mean , can you explain.