Answer: 125

Solution:

The three ways to pair the four players are equally likely, so Carl plays Azar in the semifinal with probability $\frac{1}{3}.$ In that case Carl beats Azar with probability $\frac{1}{3}$ and then beats the Jon–Sergey winner with probability $\frac{3}{4},$ so Carl wins the tournament with probability $\frac{1}{3} \cdot \frac{3}{4} = \frac{1}{4}.$

Otherwise (probability $\frac{2}{3}$) Carl plays Jon or Sergey and wins with probability $\frac{3}{4}.$ His opponent in the final is Azar with probability $\frac{3}{4}$ (Carl then wins with probability $\frac{1}{3}$) and is Jon or Sergey with probability $\frac{1}{4}$ (Carl then wins with probability $\frac{3}{4}$). So in this case Carl wins the tournament with probability $$\begin{aligned} &\frac{3}{4}\left(\frac{3}{4} \cdot \frac{1}{3} + \frac{1}{4} \cdot \frac{3}{4}\right) \\ &= \frac{3}{4} \cdot \frac{7}{16} \\ &= \frac{21}{64}. \end{aligned}$$

The total probability is $\frac{1}{3} \cdot \frac{1}{4} + \frac{2}{3} \cdot \frac{21}{64} = \frac{1}{12} + \frac{7}{32} = \frac{29}{96},$ so $p + q = 29 + 96 = 125.$