The gradient of AB is \(1/2\), so the line AB through A is \[y-3=\frac12(x-8),\qquad y=\frac12x-1.\] Its intersection with the perpendicular bisector is the midpoint M: \[\frac12x-1=-2x+4\quad\Longrightarrow\quad M=(2,0).\] Thus \[\frac{8+p}{2}=2,\quad\frac{3+q}{2}=0,\qquad\boxed{p=-4,\quad q=-3}.\]