Mind your rows and columns .....

For $3 \le k \le 6$ choose $k$ rows in $\binom{8}{k}$ ways and $k$ columns in $\binom{8}{k}$ ways. The squares where these $k$ rows and $k$ columns intersect are where we can put the $k$ pawns. In the first column from the left, choose one of the rows in one of $k$ ways. In the next column to the right, choose one of the rows in one of $k - 1$ ways. Repeat this process through to the $k$ th chosen column, for which there will be just $1$ row to choose from.

There are then $k! * \dbinom{8}{k}^{2}$ ways of placing the $k$ pawns so that no two pawns lie in the same row or column as each other.

Without restrictions, there are $\dbinom{64}{k}$ ways of placing $k$ pawns on the board. Thus

$p(k) = \dfrac{k! * \dbinom{8}{k}^{2}}{\dbinom{64}{k}}$ .

Using this formula we find that $p(3) + p(4) + p(5) + p(6) = \dfrac{232148}{334707}$ .

This gives us $a = 232148, b = 334707$ and $b - a = \boxed{102559}$ .