For $n$ leaves we have $n-1$ keys in internal node. (see 'part a' of this question)

Total keys in internal nodes $= n-1,$ each node can have keys between d and 2d.

For $n-1$ keys there will be minimum $\left \lceil \frac{n-1}{2d} \right \rceil$ internal nodes, and maximum $\left \lceil \frac{n-1}{d} \right \rceil$ internal nodes.

To calculate Big-Omega I am taking maximum everywhere.

If every node contains $d+1$ pointers (d keys) then height will be maximum, because number of nodes to be accommodated are fixed $\left (\left \lceil \frac{n-1}{d} \right \rceil \right )$.

If height is $h$ then equation becomes

$\large 1+(d+1)+(d+1)^2+(d+1)^3 +\ldots+(d+1)^{h-1} = \frac{n-1}{d}$

$\implies \frac{(d+1)^h-1}{(d+1)-1} = \frac{n-1}{d}$

$\implies (d+1)^h = n$

$\implies h =\log_{(d+1)}n$

This is the maximum height possible or say maximum number of levels possible.

Now using $h$ traverse we can get to leaf node, and then we may need to traverse 'd' more keys if our desired record is present in rightmost of leaf.

Answer is $O(h+d)$ i.e., $O(\log_{(d+1)}n+d)\\=O(\log_dn+d)\\$