Given that the number of subsets of set $A$ is 56 more than that of set $B$, we have:
$2^m - 2^n = 56$
$2^n(2^{m-n} - 1) = 56$
$2^n(2^{m-n} - 1) = 2^3 \times 7$
Comparing both sides, we have:
$2^n = 2^3$ and $2^{m-n} - 1 = 7$
$n = 3$ and $2^{m-n} = 8$
$n = 3$ and $2^{m-n} = 2^3$
$n = 3$ and $m-n = 3$
$n = 3$ and $m = 6$
So the coordinates of point P are $(6,3)$.
The coordinates of point Q are $(-2,-3)$.
The distance between $P$ and $Q$ is:
$PQ = \sqrt{(6 - (-2))^2 + (3 - (-3))^2}$
$PQ = \sqrt{(6+2)^2 + (3+3)^2}$
$PQ = \sqrt{8^2 + 6^2}$
$PQ = \sqrt{64 + 36}$
$PQ = \sqrt{100}$
$PQ = 10$
Therefore, the distance between $P(6,3)$ and $Q(-2,-3)$ is 10.