Solution:

Construct $\overline{AC}$, $\overline{AQ}$, $\overline{BQ}$, $\overline{BD}$, and let $R$ denote the intersection of $\overline{AC}$ and $\overline{BD}$. Because $ABCD$ is cyclic, we have that $\triangle ABR \sim \triangle DCR$ and $\triangle ADR \sim \triangle BCR$. Thus, we may write $AR = 4x$, $BR = 2x$, $CR = 6x$, $DR = 12x$. Now, Ptolemy applied to $ABCD$ yields $140x^{2} = 1 \cdot 3 + 2 \cdot 4 = 11$.

Now $\overline{BQ}$ is a median in triangle $ABD$. Hence,
$$BQ^{2} = \frac{2BA^{2} + 2BD^{2} - AD^{2}}{4}.$$
Likewise,
$$CQ^{2} = \frac{2CA^{2} + 2CD^{2} - DA^{2}}{4}.$$
But $PQ$ is a median in triangle $BQC$, so
$$PQ^{2} = \frac{2BQ^{2} + 2CQ^{2} - BC^{2}}{4} = \frac{AB^{2} + BD^{2} + CD^{2} + CA^{2} - BC^{2} - AD^{2}}{4} = \frac{(196 + 100)x^{2} + 1^{2} + 3^{2} - 2^{2} - 4^{2}}{4} = \frac{148x^{2} - 5}{2} = \frac{148 \cdot \frac{11}{140} - 5}{2} = \frac{116}{35}.$$

Another solution is possible. Extend $\overline{AD}$ and $\overline{BC}$ past $A$ and $B$ to their intersection $S$. Use similar triangles $SAB$ and $SCD$, and similar triangles $SAC$ and $SBD$ to compute $SA$ and $SB$, then apply the Law of Cosines twice, first to compute the cosine of $\angle A$ and then to compute $PQ^{2}$.