When the switch is at position $2,$ it is connected to $V_{cc}$ thus, the value of control input $M = 1$ which is fed to XOR gates as well. So
- $B_1 \oplus 1 = \overline{B_1}$ and
- $B_0 \oplus 1 = \overline{B_0}$
And the basic hardware is of adder only.
Now $C_{in}$ is connected to $V_{cc}$ as well. Hence, $C_{in} = 1.$
Net operation can be denoted as : $S_0 = A_0 + \overline{B_0} + 1$ and similarly for $S_1$ as well.
This is nothing but an expression of subtraction using $2's$ complement. $($Had it been $A_0 + \overline{B_0},$ it would have been addition of $1's$ complement merely, but for subtraction we need to have $2's$ complement of other operand which is $B$ here$)$
Hence, the correct answer should be : subtraction, 2's complement