Since, every adjacency matrix is also a symmetric matrix and so $A=A^T=A^{-1}$ implies $A^2=AA^T=A^TA=I_n.$
So $A_{n \times n}$ is an orthogonal matrix because both the row and column spaces of this matrix form an orthonormal basis of $\mathbb{R}^n$
(each row and column should be mutually perpendicular).
It simply means that $a_{ii}^{th}$ entry in $A^2$ is the dot/inner product of $i^{th}$ row and $i^{th}$ column. For example, $a_{11}$ in $A^2$ is the dot product of first row and first column.
Here, graph is simple and so $A$ is a matrix of 0s and 1s. Since the result of dot product is 1, it means there should be only one entry of 1 in each row and each column with the same position so that $v^Tv=\Sigma_{i=1}^nv_{i}^2=1$ for each row/column vector $v$ of 0s and 1s in $A.$ It simply means if $k^{th}$ entry of $1^{st}$ row is $1$ then the $k^{th}$ entry should also be $1$ in the $1^{st}$ column.
So, here, we can say that if $v_i$ is connected to $v_j$ then $v_j$ should also be connected to $v_i$ and there should not be any other connection for both $v_i$ and $v_j.$ In this way we can form the set of independent edges that cover the whole graph and So, $G$ will definitely have a perfect matching.
So, Answer is B.
Counter-example for other options can be $K_3.$
One more way to prove the statement is using the fact that every $(i,j)^{th}$ entry of $A^k$ is the number of $v_i,v_j-walks$ of length $k$ in $G.$
