Why the hint is required depends on the way you derived your result, not just on the result you post.
However, here's an alternative solution.
Denote
$$ p_{k,n}:=P(\text{Player #$k$ wins an $n$ round tournament}).$$
As a preliminary step, we should eliminate the cases $p=\frac12$ as then everything is symmetric, i.e. the probability that player #2 (or anybody else) wins, is exactly $\frac1n$. Similarly, if $p=1$, then #1 wins the tournament a.s., hence $p_{2,n}=0$. And if $p=0$, then the max player will win a.s., i.e. then $p_{2,n}=0$ except that $p_{2,1}=1$.
After these remarks, we don't get into trouble if we wish to divide by $p$, or $1-p$ or $1-2p$ below.
Next, let us compute the probability that player $1$ wins the tournament. To do so, he must survive the first round and then win a tournament of $2^{n-1}$ players among which he (after renumbering the players to fill the gaps without changing relative order) is still number one. Then we have
$$ p_{1,n}=\begin{cases}1&\text{if }n=0,\\
p\cdot p_{1,n-1}&\text{if }n>0,\end{cases}$$
In other words
$$p_{1,n}=p^n.$$
Now what about player $2$? To win, he must survive the first round; that is he
- either plays player #1 (with probaility $\frac1{2^n-1}$) and wins with probaility $1-p$, after which he is automatically the #1 of the remaining players
- or plays somebody else (with $1-\frac1{2^n-1}$) and wins with probability $p$. Player #1 also wins with porbability $p$, and then player #2 is still #2 of the remaining players
- or as before, but player #1 loses his match, so our #2 becomes #1 of the survivors.
That is, for $n>0$ (otherwise, a player #2 doesn't even participate) we have
$$\begin{align}p_{2,n}&=\frac1{2^n-1}\cdot(1-p)\cdot p_{1,n-1}+\left(1-\frac1{2^n-1}\right)\cdot\left( p^2\cdot p_{2,n-1}+ p\cdot (1-p)\cdot p_{1,n-1}\right)
\\&=\frac{p^{n-1}(1-p)}{2^{n}-1} +\frac{(2^{n}-2)(1-p)p^n}{2^{n}-1}+\frac{(2^{n}-2)p^2}{2^{n}-1}\cdot p_{2,n-1}.
\end{align}$$
(Note that this immediately gives $p_{2,1}=1-p$).
If we substitue $p_{2,n}=\frac{p^n}{2^n-1}(a_n+c)$, we obtain
$$ a_n +c= \frac{1-p}{p}+(2^n-2)(1-p)+2p(a_{n-1}+c),$$
hence by chosing $c=\frac{1-p}{p(1-2p)}$
$$ a_n = (2^n-2)(1-p)+2pa_{n-1}.$$
Note that $p_{2,1}=1-p$ implies $a_1=\frac{2(1-p)}{2p-1}$, so we can extend this to $a_0=\frac{1-p}{(2p-1)p}$.
Substitute again, $a_n=2^np^nb_n$, to find
$$ b_n = \frac{(2^n-2)(1-p)}{2^np^n}+b_{n-1}= (1-2^{1-n})\frac{(1-p)}{p^n}+b_{n-1}.$$
Hence
$$\begin{align} b_n &= b_0+\sum_{k=1}^n(1-2^{1-k})\frac{(1-p)}{p^k}\\&=b_0 + (1-p)\sum_{k=1}^np^{-k}-2(1-p)\sum_{k=1}^n(2p)^{-k}\\&
= b_0 + (p^{-n}-1) -2(1-p) \frac{(2p)^{-n}-1}{1-2p}\\
&= \frac1{p^{n}}-\frac{2(1-p)}{(1-2p)(2p)^n} +\left(b_0-1+\frac{2(1-p)}{1-2p}\right)
\\&=\frac1{p^{n}}-\frac{2(1-p)}{(1-2p)(2p)^n}-\frac1p\end{align}$$
Therefore,
$$ a_n = 2^n-\frac{2(1-p)}{1-2p}-2^np^{n-1}$$
and ultimately,
$$ p_{2,n} = \left(2^n-2^np^{n-1}+\frac{1-p}{p}\right)\frac{p^n}{2^n-1}.$$
There isn't enough information to answer the question, since we don't know the probabilities of the players winning their games. I'll assume that the outcomes of all games are independent and either player has a $50\%$ chance of winning each game.
The probability that $S_2$ will win is $\frac14$. If she doesn't, every other player has the same chance.
Edit:
Here's a more explicit solution. The probability that $S_2$ will win is $\frac14$, since she reached the semifinal. Thus the probability that someone else will win is $\frac34$. No-one else is special, so they all have the same probability to win. Since there are $2^n-1$ of them, the probability that one of them, $S_1$, wins is $\frac34\left(2^n-1\right)^{-1}$. Equating this to $\frac1{20}$ then yields $n$.
Best Answer
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 $$ \frac{\binom{12}{3}}{\binom{15}{3}} = \frac{44}{91} \approx 0.48. $$