For master theorem, recurrence should be of the form $T(n)=aT(n/b)+f(n)$, where $a\geq 1$ and $b> 1.$. Intuitively, the 3 cases of the theorem are:
Case(1): $n^{\log_b(a)}$ should be polynomially larger than $f(n)$
Case(2): $n^{\log_b(a)}$ should be equal to $f(n)$
Case(3): $n^{\log_b(a)}$ should be polynomially smaller than $f(n)$
So, in option(A): $n^{\log_b(a)}=n$ and $f(n)=n\log n,$, which seems to be case (3). But, here $f(n)/n^{\log_ba})=\log n,$ which is not polynomial larger. So Cannot apply Master theorem. [Cormen, Page 96-97]. But we can use Extended Master theorem here and get $T(n) = \Theta\left(n \log^2 n \right)$ as shown here.
Option (b) is case (2), so $T(n)=\Theta(n)$
Option(c) is case(1), so $T(n)=\Theta(n^3)$
Option(d) is case(3), so $T(n)=\Theta(n^2)$. Case 3 also requires that $af(n/b) \leq cf(n)$ for some $c < 1$ and all sufficiently large $n$. Here, we get $7 (n/4)^2 \leq c n^2$ is true for $c \geq 7/16$ and all $n\geq 1.$
The Answer is Option A
