Solution:

The answer is (A). Let us number the rows from $1$ to $8$ and the columns from $1$ to $8$; consider the central $2 \times 2$ square, the one formed by columns $4$ and $5$ intersected with rows $4$ and $5$. We examine two cases separately:

a. In the square we place an $A$ and a $B$ in cells that meet at a vertex, and a $C$ and a $D$ in the two remaining cells; we can do this in $8$ different ways. Once the square has been filled in one of these ways, there is one and only one way to complete the entire chessboard.

b. The $A$ and the $B$ placed in the central square are horizontally or vertically adjacent (we can fill the square in $4! - 8$ ways so that this happens). Suppose that $A$ is located in row $4$, column $4$ and $B$ in row $4$, column $5$ (the other cases are perfectly analogous). Row $4$ must be filled with $A$ and $B$ alternating; row $5$ must in turn contain $C$ and $D$ alternating (starting with $C$ or with $D$ depending on how we filled the central square). The odd rows must all be identical to row $5$. The even rows will contain $A$ and $B$ alternating: in row $4$ the starting letter is fixed (in our case it is $A$) but for all the other even rows ($2$, $6$ and $8$) we can choose (independently) whether they start with $A$ or with $B$. Altogether we have counted $(4! - 8) 2^3$ configurations.

In total the admissible configurations are therefore $136$.