If the Cayley table is symmetric along the diagonal axis, then the $(ij)^{th}$ entry is equal to the $(ji)^{th}$ entry, so $a_i\ast a_j=a_j\ast a_i$, and so the group is abelian.
Conversely, if the group is abelian, then $a_i\ast a_j=a_j\ast a_i$, for every $a_ia_j$ in our group. Therefore, the $(ij)^{th}$ entry is equal to the $(ji)^{th}$ entry in the Cayley table, so the table is symmetric.
$$\require{color}
\begin{array}{| c | c| c | c | c |}
\hline
\textbf{$\ast$} & \textbf{a} &\textbf{b} &\textbf{c} & \textbf{d} \\ \hline
\textbf{a}& d & c & \textcolor{red}{b} & a \\ \hline
\textbf{b}& \textcolor{blue!90!black}{c} & d & a & \textcolor{red}{b} \\ \hline
\textbf{c}& b & \textcolor{magenta!70!black}{a} & d & c \\ \hline
\textbf{d} & \textcolor{magenta!70!black}{a} & b & \textcolor{blue!90!black}{c} & d \\
\hline
\end{array}$$There is only one way to fill the Cayley table as one of the symmetric entries are already filled.
So, the correct answer is $(C)$.
