jmc

Solution 1: The segment $BD$ begs to be drawn, so we start there: We can clearly see that now we have triangles $ABD$ and $CBD,$ and $MD,$ $ND,$ and $AC$ are medians to one or more of the triangles. That means that $P$ and $Q$ are the centroids of triangles $ABD$ and $CBD,$ respectively. Since $AC=15,$ that means $CQ=5,$ since the median from $C$ to $BD$ is half the length of $AC,$ or $7.5,$ and $CQ$ must be $\frac{2}{3}$ of that, or $5.$ Therefore, $QA=AC-CQ=15-5=\boxed{10}.$

Solution 2: Since $ABCD$ is a parallelogram, $\overline{AD}$ and $\overline{BC}$ are parallel with $\overline{AC}$ and $\overline{DN}$ as transversals. So $\angle DAQ=\angle NCQ$ and $\angle ADQ=\angle CNQ$, and so $△ADQ$ and $△CNQ$ are similar by AA similarity.

Also, we know opposite sides of a parallelogram are congruent, so $AD=BC$. Since $N$ is a midpoint of $\overline{BC}$, we have $CN=\frac{AD}{2}$. By similar triangles, $$\frac{AQ}{CQ}=\frac{AD}{CN}=2,$$so $AQ=2CQ.$ Since $AQ+CQ=AC=15$, we have $CQ=5$ and $AQ=\boxed{10}.$