What is the probability that the player $P_4$ reaches the semi final

Player $P_4$ shares their semi-final group with three other players out of the remaining 15; they advance to the semi-final if and only if none of those three players are $P_1, P_2, P_3$. This means the probability of them advancing is $$ \frac{\binom{12}{3}}{\binom{15}{3}} = \frac{44}{91} \approx 0.48. $$

In the first round $P_4$ plays with equal probability against any other team. The probability that $P_4$ wins this round is therefore given by ${12\over15}$. We now condition on this event, i.e., that in the first round $P_4$ has played against a $P_k$ with $k>4$.

In the second round there are $7$ adversaries left for $P_4$, all of them equiprobable. We have now have to see how many of them are better than $P_4$. This depends on whether in the first round there was a match among $1$–$3$. Denote the probability for this to have happened by $p$. Given that $P_4$ in the first round has not played against one of $1$–$3$ we obtain $$p={2\over13}+{11\over13}\cdot{1\over11}={3\over13}\ .\tag{*}$$ In the case covered by $p$ two of the adversaries of $P_4$ are better than $P_4$, in the case covered by $1-p$ there are three of them. It follows that the overall probability $p_*$ that $P_4$ wins in both rounds is given by $$p_*={4\over5}\bigl(p\cdot{5\over7}+(1-p){4\over7}\bigr)={44\over91}=0.484\ .$$ $$$$ $(^*)$ The probability that $1$ plays against $2$ or $3$ is ${2\over13}$. If $1$ plays against a $P_k$ with $k>4$ the probability that $2$ plays against $3$ is ${1\over11}$.

$P_4$ will reach the semifinal if and only if none of $P_1,P_2,P_3$ is in the same quarter of the final draw as $P_4$.

It follows that the probability that $P_4$ reaches the semifinal is $${\small{\frac{\binom{12}{3}}{\binom{15}{3}}}}={\small{\frac{44}{91}}}$$ $$ \overline{ \qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\;\;\; }$$ The above solution assumes the distribution of the final draw, assuming random repairings, is the same as the distribution of a random draw, with no repairings. That seemed intuitively clear (by symmetry), but based on lulu's comment, I began to worry about whether the random repairings were perhaps relevant, hence I tried a different approach, as shown below, which explicitly assumes random repairings.

With some corrections applied, based as on comments by Henry and Mees de Vries, here is a revised version of that approach . . .

The probability that $P_4$ gets to round two is ${\large{\frac{12}{15}}}={\large{\frac{4}{5}}}$, since in round one, $P_4$ has to avoid playing any of $P_1,P_2,P_3$.

Consider two cases . . .

Case $(1)$:$\;$In the first round, none of $P_1,P_2,P_3$ plays $P_4$, and two of $P_1,P_2,P_3$ play each other.

The probability that this case occurs is $$ \left({\small{\frac{4}{5}}}\right)\left({\small{\frac{3}{13}}}\right) = {\small{\frac{12}{65}}} $$ Given that this case occurs, the probability that $P_4$ wins the next match is ${\large{\frac{5}{7}}}$, since in round two, $P_4$ needs to avoid playing any of the two remaining stronger players.

Case $(2)$:$\;$In the first round, none of $P_1,P_2,P_3$ plays $P_4$, and no two of $P_1,P_2,P_3$ play each other.

The probability that this case occurs is $$ \left({\small{\frac{4}{5}}}\right)\left({\small{\frac{10}{13}}}\right) = {\small{\frac{8}{13}}} $$ Given that this case occurs, the probability that $P_4$ wins the next match is ${\large{\frac{4}{7}}}$, since in round two, $P_4$ needs to avoid playing any of $P_1,P_2,P_3$.

Summing the results for the two cases, it follows that the probability that $P_4$ reaches the semifinal is $$ \left({\small{\frac{12}{65}}}\right)\left({\small{\frac{5}{7}}}\right) + \left({\small{\frac{8}{13}}}\right)\left({\small{\frac{4}{7}}}\right) = {\small{\frac{44}{91}}} $$ Note:$\;$The two approaches yield the same result, as I initially expected, but wasn't sure of.