There are two control line one is $K$ and another is $C_{0}$ .

When $K=1$ , $C_{0}= 1$ we can perform $A-B$

When $K=0$, $C_{0}=0$ we can perform $A+B$

But without manipulating $B(B_{0},B_{1},\ldots)$ we cannot perform A+1. But here we have only two control lines which is $K,C_{0}$. Therefore the answer is A.

Note:

**For A+B, $C_{0}= 0$ , $K=1$, and $0 \oplus x = x$**

$\frac{\begin{matrix} &A3 &A2 &A1 &A0 \\ +& B3 & B2 & B1 & B0 \end{matrix}}{\begin{matrix} &S3&S2&S1&S0\end{matrix}}$

$\frac{\begin{matrix} &C3&C2&C1&C0=0\\ &A3 &A2 &A1 &A0 \\ \qquad\qquad\qquad\qquad+& B3 & B2 & B1 &B0 \end{matrix}}{\begin{matrix} Sum \enspace Output:&S3&S2&S1&S0\\Carry \enspace Output:\qquad&C4&C3&C2&C1\end{matrix}}$

**For A- B, $C_{0} = 1$ , $K = 1$, and $1\oplus x=\bar x$**

Subtraction Using 1's complement, $(A -B)$

1. Find 1's complement of $B$= $2^{n}-1 -B$, We do 1's complement but change bits 0 to 1 or 1 to 0 (where n is no of bits)

2. Add A , $2^{n} -1 - B +A = 2^{n}-1+(A-B)$ but we need only A-B , $2^{n}$ is last carry , That is, $C_{4}$ will be discarded, to remove -1, add $C_{0}=1$.

$\frac{\begin{matrix} &A3 &A2 &A1 &A0 \\ -& B3 & B2 & B1 & B0 \end{matrix}}{\begin{matrix} &S3&S2&S1&S0\end{matrix}}$

Subtraction using 1's complement

$\frac{\begin{matrix} &C3&C2&C1&C0=1\\ &A3 &A2 &A1 &A0 \\ \qquad\qquad\qquad\qquad+& \overline{B3} & \overline{B2} & \overline{B1} &\overline{B0} \end{matrix}}{\begin{matrix} Sum \enspace Output:&S3&S2&S1&S0\\Carry \enspace Output:\qquad&C4&C3&C2&C1\end{matrix}}$