Lower bound for $k$-sorting an array
From the result of $k$-sorting an array you can tell which elements of the original array were the $n/k$ smallest, the $n/k$ largest, and which belong to each of the other $k-2$ blocks of $n/k$ elements in between. In particular, there are at least $\binom{n}{n/k,\ldots,n/k}$ different leaves, which works out to $\frac{n!}{(n/k)!^k} = \exp \Theta(n\log k)$. This gives you the correct lower bound.

What you calculated is the number of different ways to completely sorted a $k$-sorted permutation. Therefore your lower bound is for sorting a $k$-sorted permutation, and for this task your lower bound is tight: a $k$-sorted permutation can be sorted in $O(n\log (n/k))$.