Solution:
Set $BC = a$, $CA = b$ and $AB = c$. Then $BA \cdot BM = BP \cdot BQ$, and
$$\frac{BM}{c} = \frac{BP}{a}$$
Hence $BQ = \frac{c^2}{a}$ and analogously $CP = \frac{b^2}{a}$. Then $BP = \frac{a^2 - b^2}{a}$, $CQ = \frac{a^2 - c^2}{a}$ and the condition $\frac{BP}{CQ} = \frac{AB}{AC}$ becomes

$$b\left(a^2 - b^2\right) = c\left(a^2 - c^2\right) \text{, i.e. } (b-c)\left(a^2 - b^2 - c^2 - bc\right) = 0$$
Since $b \neq c$, we get $a^2 - b^2 - c^2 - bc = 0$ and the Cosine theorem gives $\text{BAC = - 1 2}$. Therefore $\text{BAC = 120}$.

Second solution. Since the quadrilateral $AMPN$ is a cyclic trapezoid, it follows that $AM = NP$. Also, if $T = MP \cap NQ$, then $AMTN$ is a parallelogram and $AM = NT$. Then $NP = NT$ and in the same way $MQ = MT$. Hence $TPN$ and $TQM$ are similar isosceles triangles and we have
$$\frac{TP}{TQ} = \frac{TN}{TM}$$
Using the Sine theorem, we obtain $\frac{MP}{\sin \beta} = \frac{BP}{\sin \alpha}$, $\frac{NQ}{\sin \gamma} = \frac{CQ}{\sin \alpha}$ and since $\frac{BP}{CQ} = \frac{AB}{AC} = \frac{\sin \gamma}{\sin \beta}$, we conclude that $MP = NQ$. From here and (1) it follows that
$$TM + TQ \frac{TN}{TM} = TN + TQ \Longleftrightarrow (TM - TN)(TM - TQ) = 0$$
Assume that $TM = TN$. We have $MQ = NP$ and $NA = MA$. The first of these identities shows that $MN \parallel PQ$ and by the second one we obtain $AC = AB$, a contradiction.
Therefore $TM = TQ$, i.e. $\triangle MTQ$ is equilateral. Hence
$$\text{BAC = MTN = 120}$$