Answer is B.
For binary search on $n$ elements we pick a path of $n/2$ elements after doing a single comparison $(O(1))$ and discard the remaining $n/2$ elements So, $T(n) = T(n/2) + 1.$
For merge sort on $n$ elements we divide the array into two equal parts containing $n/2$ elements, recursively merge sort them and finally merges the $2$ arrays in $O(n)$ time. So, $T(n) = 2T(n/2) + cn.$
For quick sort on $n$ elements we split the array based on the position of the pivot. If pivot happens to be the $k+1^{th}$ element in the sorted array, we do a recursive call using $k$ elements and $n-k-1$ elements and we need $O(n)$ time to decide the position of the pivot. So, $T(n) = T(n-k-1) + T(k) + cn.$ (A $“-1”$ is missing in the question option).
For Tower of Hanoi using $n$ disks, to position the first disk we have to replace $n-1$ disks two times. To place the remaining disks we have to recursively solve the problem thus giving $T(n) = 2T(n-1) + 1.$