Number of elements in the path from new leaf to root $= \log n$, and all elements are sorted from leaf to root so we can do a binary search which will result in $O(\log \log n)$ number of comparisons.
Since a heap is a complete binary tree (hence height balanced also), in an array implementation, from every node index, we can know its depth and this will be the number of elements $-$ $n$ for binary search.
Correct Answer: $B$