Answer: 242

Small Hint:

Let the bisector of $\angle A$ meet $\overline{CD}$ at $A':$ alternate interior angles make triangle $ADA'$ isosceles with $DA' = 333$

Big Hint:

In isosceles triangle $ADA',$ the bisector from $D$ is also the median, so $P$ is the midpoint of $AA'.$ Do the same for $Q$ and use coordinates.

Solution:

Let the bisector of $\angle A$ meet $\overline{CD}$ at $A'.$ Since $\overline{AB} \parallel \overline{CD},$ we have $\angle DA'A = \angle A'AB = \angle A'AD,$ so triangle $ADA'$ is isosceles with $DA' = DA = 333.$ The bisector of $\angle D$ is then the median from $D$ in this triangle, so $P,$ which lies on both bisectors, is the midpoint of $\overline{AA'}.$ Symmetrically, $Q$ is the midpoint of $\overline{BB'},$ where $B'$ is on $\overline{CD}$ with $CB' = 333.$

Place $D = (0, 0)$ and $C = (650, 0),$ so $A = (75, h)$ and $B = (575, h)$ for the appropriate height $h.$ Then $A' = (333, 0)$ and $B' = (650 - 333, 0) = (317, 0),$ so $$\begin{aligned} P &= \left(\frac{75 + 333}{2}, \frac{h}{2}\right) \\ &= \left(204, \frac{h}{2}\right), \end{aligned}$$ $$\begin{aligned} Q &= \left(\frac{575 + 317}{2}, \frac{h}{2}\right) \\ &= \left(446, \frac{h}{2}\right). \end{aligned}$$

Therefore $PQ = 446 - 204 = 242.$