C is the correct answer.
$1 << d$ returns a number with ${(d+1)}^{th}$ bit from right side set to $1.$ So, bitwise OR with this will set the ${(d+1)}^{th}$ bit and bitwise XOR will complement it. But bitwise AND will also affect the other bits and it is not doing the unset operation.
if $n$ is a power of $2$ it'll have just one $“1"$ and all bits to its right will be $0's.$ And $n-1$ will have $1's$ in all those bits corresponding to the $0's$ in $n.$ So, $n\&(n-1)$ returns $0$ for any power of $2.$