This question is dealing with all the 3 cases of Euler's totient function
$\phi (n)$ = n - 1 when n is a prime number.
$\phi (n)$ = $\phi (p)$$\phi (q)$ if n = p.q & both p and q are prime numbers
$\phi (n)$ = $p^{k} - p^{k-1}$ if n = $p^{k}$ where p is prime.
now, $\phi (n)$ = $\phi (p^{2}.q)$ =$\phi (p^{2}).\phi (q)$ = $(p^{2} - p^{1}).(q-1)$ = p.(p-1).(q-1)
option D is correct