Solution:

a. Let $K$ be the point of intersection between $AP$ and $BC$ (that is, the foot of the altitude issuing from $A$). Since $\widehat{AKB} = 90^\circ$, we have that $\widehat{BAK} = 90^\circ - \widehat{ABC}$. Moreover $\widehat{ABC} = \frac{1}{2} \widehat{AOC}$ (angles respectively at the circumference and at the center subtending the same arc). Finally, since $AOC$ is isosceles ($AO$ and $CO$ are radii of the circumcircle of $ABC$), we also have that $\widehat{AOC} = 180^\circ - 2\widehat{OAC}$. Therefore
$$\widehat{BAP} = \widehat{BAK} = 90^\circ - \widehat{ABC} = 90^\circ - \frac{1}{2} \widehat{AOC} = 90^\circ - \frac{1}{2}(180^\circ - 2\widehat{OAC}) = \widehat{OAC} = \widehat{QAC}$$

b. $\widehat{BCP} = \widehat{BAP}$ since they are angles at the circumference subtending the same arc. For the same reason, $\widehat{QBC} = \widehat{QAC}$. Then, by what was proved in point (a), $\widehat{BCP} = \widehat{BAP} = \widehat{QAC} = \widehat{QBC}$. Moreover $\widehat{BPC} = \widehat{BQC}$ (again because they are angles at the circumference subtending the same arc).
Let us now consider the triangles $BCP$ and $CBQ$. We have shown that they have two pairs of equal angles, but then by subtraction the third angle is also equal in the two triangles: $\widehat{CBP} = \widehat{BCQ}$. These triangles also have the side $BC$ in common, hence they are congruent by the second congruence criterion.

c. By the congruence proved in point (b), the area of $ABPC$ is equal to the area of $ABQC$. Moreover, by definition of $M, N$ and $O$, we have that $AB = 2AM$, $AC = 2AN$ and $AQ = 2AO$; hence the homothety of center $A$ and factor $2$ sends the quadrilateral $AMON$ to the quadrilateral $ABQC$. The ratio between the area of $ABQC$ and the area of $AMON$ is therefore $2^2 = 4$.