There are three ways for Team $A$ to win the playoff: win the first two games; win the first game, lose the second game, and win the third game; or lose the first game and win the second and third games. The probability that it wins in one of these ways is
$$\frac{2}{3} \cdot p + \frac{2}{3} \cdot (1-p) \cdot p + \frac{1}{3} \cdot p^2 = -\frac{1}{3}p^2 + \frac{4}{3}p.$$
Setting this equal to $\frac{1}{2}$ and simplifying gives $2p^2 - 8p + 3 = 0$, and the Quadratic Formula gives solutions $\frac{1}{2}(4 \pm \sqrt{10})$. Choosing the plus sign gives a nonsensical value of $p$ because it is greater than $1$, so the required probability is $\frac{1}{2}(4 - \sqrt{10}) \approx 0.42$. The requested sum is $4 + 10 = 14$.