And so we may conclude the 3 ratios are an arithmetic progression. This follows from the $x$-coordinates of points $B$, $M$ and $C$ being an arithmetic progression.
This is a generalisation of the problem in this thread, and one way to tackle it is by using vector geometry, as Pranav did in that thread.
Let $\lambda = \dfrac{{AB}}{{AP}}$, $\mu = \dfrac{{AM}}{{AN}}$, $\nu = \dfrac{{AC}}{{AQ}}$, $\mathbf{b} = \vec{AB}$ and $\mathbf{c} = \vec{AC}$. Then $$\vec{AP} = \frac1\lambda \mathbf{b},\quad \vec{AN} = \frac1{2\mu}(\mathbf{b} + \mathbf{c}), \quad \vec{AQ} = \frac1\nu\mathbf{c}.$$ The points $P,\ N,\ Q$ are collinear, so $\vec{PN} = \alpha\,\vec{QN}$ for some scalar $\alpha$. Hence $$\frac1{2\mu}(\mathbf{b} + \mathbf{c}) - \frac1\lambda \mathbf{b} = \alpha\Bigl(\frac1{2\mu}(\mathbf{b} + \mathbf{c}) - \frac1\nu\mathbf{c} \Bigr).$$ Compare coefficients of $\mathbf{b}$ and $\mathbf{c}$ to get $$\frac1{2\mu} - \frac1\lambda = \frac{\alpha}{2\mu}, \qquad \frac1{2\mu} = \frac\alpha{2\mu} - \frac{\alpha}{\nu}.$$ Thus $\dfrac1\lambda = -\dfrac{\alpha}{\nu}$ so that $\alpha = -\dfrac\nu\lambda.$ Substitute that into the previous displayed equation to get $$\frac1{2\mu} - \frac1\lambda = -\frac{\nu}{2\lambda\mu}.$$ Clearing fractions, you see that $\lambda - 2\mu + \nu = 0$, which means that $\lambda$, $\mu$ and $\nu$ are in arithmetic progression.