Inspired by Roopesh

$P_{n} =\displaystyle \prod_{k=0}^{n}\binom{n}{k}$

$= \displaystyle \prod_{k=0}^{n} \dfrac{n!}{(n-k)!\cdot k!}$

$= n!^{n+1} \displaystyle \prod_{k=0}^{n} \dfrac{1}{k!^{2}}$

$\therefore P_{n+1} = (n+1)!^{n+2} \displaystyle \prod_{k=0}^{n+1} \dfrac{1}{k!^{2}}$

$\Rightarrow \dfrac{P_{n+1}}{P_{n}}=\dfrac{(n+1)^{n}}{n!} ,\dfrac{P_{n}}{P_{n-1}}=\dfrac{(n)^{n-1}}{(n-1)!}$

Now the question asks for,

$\displaystyle \lim_{n\rightarrow \infty} \dfrac{P_{n-1}P_{n+1}}{P_{n}^{2}}$

So we have ,

$\displaystyle \lim_{n\rightarrow \infty} \dfrac{P_{n-1}P_{n+1}}{P_{n}^{2}} = \lim_{n\rightarrow \infty} \dfrac{(n-1)!(n+1)^{n}}{n!\times n^{n-1}}$

$= \displaystyle \lim_{n\rightarrow \infty} \dfrac{(n+1)^{n}}{n\times n^{n-1}}$

$=\displaystyle \lim_{n\rightarrow \infty} \left ( \dfrac{n+1}{n} \right )^{n}$

$= \displaystyle \lim_{n\rightarrow \infty} \left ( 1 + \frac{1}{n} \right )^{n}$

$= e$