Solution:

Let $p(h)$ be the probability that one player flips $h$ heads. Splitting according to the two fair coins and the biased coin, $$p(0) = \tfrac{1}{4} \cdot \tfrac{3}{7} = \tfrac{3}{28}, \quad p(1) = \tfrac{2}{4} \cdot \tfrac{3}{7} + \tfrac{1}{4} \cdot \tfrac{4}{7} = \tfrac{10}{28}, \quad p(2) = \tfrac{1}{4} \cdot \tfrac{3}{7} + \tfrac{2}{4} \cdot \tfrac{4}{7} = \tfrac{11}{28}, \quad p(3) = \tfrac{1}{4} \cdot \tfrac{4}{7} = \tfrac{4}{28}.$$

Jackie's and Phil's flips are independent with the same distribution, so the probability that their head counts agree is $$\sum_h p(h)^2 = \frac{3^2 + 10^2 + 11^2 + 4^2}{28^2} = \frac{246}{784} = \frac{123}{392}.$$ Thus $m + n = 123 + 392 = 515.$