This time it's correct!

If $p$ is any 3-digit prime,then $p^{2} > 2000$ .

$\implies$ highest power of $p$ that divides $2000!$ is $[\dfrac{2000}{p}]$ and $1000!$ is $[\dfrac{1000}{p}]$ .

$\implies$ highest power of $p$ that divides $\dfrac{2000!}{1000!}$ is $[\dfrac{2000}{p}]$ $- [\dfrac{1000}{p}]$ .

If we put $997 = p$ we get $[\dfrac{2000}{997}]$ $- [\dfrac{1000}{997}] = 1$

$\implies$ largest prime that divides ${ P }_{ 1000 }^{ 2000 }$ is $997$ .