In insertion sort, with linear search, it takes
(worst case) $n$ comparisons for searching the right position, and $n$ swaps to make room to place the element.
Hence for n elements, a total of $n\times(n+n)$; $n$ for search and $n$ for swaps.
$= \Theta (2n^2) = \Theta (n^2)$
If we replace it with binary search, it takes
(worst case) $\log n$ comparisons for searching the right position, and $n$ swaps to make room to place the element.
Hence for n elements, a total of $n\times(\log n+n)$; $n$ for search and $n$ for swaps.
$= \Theta (n \times \log n + n^2) = \Theta (n^2)$
Hence, answer is A.