The easiest way to solve this problem is to substitute a value of n which satisfies \(n \leq -5\)
Lets take n = -5 and put in all the option choices

A. \(\frac{1}{(n+3)^2}\)
n = -5
=> \(\frac{1}{(n+3)^2}\) = \(\frac{1}{(-5+3)^2}\) = \(\frac{1}{(-2)^2}\) = \(\frac{1}{4}\) = 0.25

B. \(- \frac{1}{(n+3)^2}\)
n = -5
=> \(- \frac{1}{(n+3)^2}\) => this value is same as option A but with negative sign. So value will be -0.25

C. \(\frac{1}{n+3}\)
n=-5
=> \(\frac{1}{n+3}\) = \(\frac{1}{-5+3}\) = \(\frac{1}{-2}\) = -\(\frac{1}{2}\) = -0.5

D. \(\frac{1}{n+4}\)
n=-5
=> \(\frac{1}{n+4}\) = \(\frac{1}{-5+4}\) = \(\frac{1}{-1}\) = -1

E. \(\frac{1}{n-4}\)
n=-5
=> \(\frac{1}{n-4}\) = \(\frac{1}{-5-4}\) = \(\frac{1}{-9}\) = -\(\frac{1}{9}\) = -0.11

Clearly, option D is the least value.
Hope it helps!