Answer is option B.
For any binary no, FSM read input from LSB and remain unchanged till first $1$, and it complement after that
$\mathbf{100} \rightarrow \mathbf{100}$ $[1$'s complement of $100 + 1 = 011+ 1 = 100 = 2$'s complement of $100]$
$0\mathbf{10} \rightarrow 1\mathbf{10}$ $[1$'s complement of $010 + 1 = 101 + 1 = 110 = 2$'s complement of $010]$
$1010\mathbf{100} \rightarrow 0101\mathbf{100}$ $[1$'s complement of $1010100 + 1 = 0101011+ 1 = 0101100]$
Note : Underline part is unchanged (till first $1$ from lsb) then $0$'s changed to $1$'s and $1$'s changed to $0$'s