Given that -
$c_1n^{n+\frac{1}{2}}e^{-n}\leq n!\leq c_2n^{n+\frac{1}{2}}e^{-n}$.
$n!=\Theta\left(\frac{n^{n+1/2}}{e^n}\right)$
We can divide it by $e^{1/2}$ (multiplying or dividing by a +ve constant doesn't affect the asymptotic nature of a function)
$\implies n!=\Theta\left(\frac{n^{n+1/2}}{e^{n+1/2}}\right)=\Theta\left(\left(\frac{n}{e}\right)^{n+\frac{1}{2}}\right)$
Option (D) is the answer.