Let circumcircle of $\bigtriangleup MPC$ intersects $AN$ again at $S$ and circumcircle of $\bigtriangleup BQN$ intersects $AM$ again at $R$. It is clear that $\angle BAM = \angle CMP = \angle ASC$ and $\angle CAN = \angle QNB = \angle ARB$. Hence, triangles $ASC$ and $RAB$ are similar and $\frac{CS}{AB} = \frac{AS}{AR}$. Using law of sines for triangles $BAM, MAN$ and $CSN$ we got the following relations. $$\frac{\sin \angle SNC}{CS} = \frac{\sin \angle NSC}{CN};$$$$\frac{\sin \angle AMN}{AN} = \frac{\sin \angle MNA}{AM};$$$$\frac{\sin \angle BAM}{BM} = \frac{\sin \angle BMA}{AB}.$$ Multiplying these relations we got that $CS \cdot AN \cdot BM = CN \cdot AM \cdot AB$. But $CN = BM$. Hence, $CS \cdot AN = AB \cdot AM$. Thus $\frac{CS\cdot AN}{AB \cdot AM} = \frac{AS \cdot AN}{AR \cdot AM}$ (we have used the fact of similarity $\bigtriangleup ASC$ and $\bigtriangleup RAB$). Hence, $AS \cdot AN = AR \cdot AM$ and it means that quadrilateral $RMNS$ is cyclic. So, $ \angle MRN = \angle MSN$. But $ \angle MRN = \angle QBN = \angle QBC$ and $\angle MSN = \angle MCP = \angle PCB$. So, $\angle QBC = \angle PCB$ as desired.