= Solution
It suffices first to prove the <elementary primorial bound> for positive integers $n$, by strong induction. The case $n=1$ is immediate.
If $n=2k$, every prime $p$ with $k<p\leq2k$ divides $\binom{2k}{k}$: the numerator $(2k)!$ contains $p$, whereas the two copies of $k!$ do not. Since these primes are distinct,
$$
P(2k)\leq P(k)\binom{2k}{k}.
$$
The induction hypothesis and the binomial theorem give
$$
P(2k)\leq4^k\binom{2k}{k}
\leq4^k2^{2k}=4^{2k}.
$$
If $n=2k+1$, part (b) shows that every prime $k+2\leq p\leq2k+1$ divides $\binom{2k+1}{k+1}$. Hence
$$
P(2k+1)\leq P(k+1)\binom{2k+1}{k+1}.
$$
The two central coefficients of $(1+1)^{2k+1}$ are equal, so
$$
2\binom{2k+1}{k+1}\leq2^{2k+1},
$$
and therefore
$$
P(2k+1)\leq4^{k+1}4^k=4^{2k+1}.
$$
This completes the induction. For real $X\geq1$,
$$
\boxed{P(X)=P(\lfloor X\rfloor)\leq4^{\lfloor X\rfloor}\leq4^X.}
$$
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