smc

Let the intersection of $\overline{AC}$ and $\overline{BD}$ be $E$. Since $ABCD$ is a rhombus, we have $\overline{AC}\perp \overline{BD}$ and $AE=CE=\frac{AC}{2}=8$. Since $\overline{NQ}\perp \overline{BD}$, we have $\overline{NQ}\parallel \overline{AC}$, so $△BNQ\sim △BAE\sim △NAP$. Therefore, $$\frac{NP}{AP}=\frac{NP}{8-NQ}=\frac{BE}{AE}=\frac{15}{8}\Rightarrow NP=15-\frac{15}{8}NQ.$$ By Pythagorean Theorem, $$P{Q}^2=N{Q}^2+N{P}^2=N{Q}^2+{\left(15-\frac{15}{8}NQ\right)}^2=\frac{289}{64}N{Q}^2-\frac{225}{4}NQ+225.$$ The minimum value of $P{Q}^2$ would give the minimum value of $PQ$, so we take the derivative (or use vertex form) to find that the minimum occurs when $NQ=\frac{225\cdot 8}{289}$ which gives $P{Q}^2=\frac{225\cdot 64}{289}$. Hence, the minimum value of $PQ$ is $\sqrt{\frac{225\cdot 64}{289}}=\frac{120}{17}$, which is closest to $7\Rightarrow \boxed{7}$.