Finding a lower bound for the amount of comparisons for sorting $k$ subarrays with $\frac n k$ elements
If a decision tree has at least $m$ leaves then its height is at least $\log_2 m$. It is the height that bounds the number of comparisons rather than the number of leaves. In your case there are $(n/k)!^k$ leaves, and so $\Omega(\log[(n/k)!^k]) = \Omega(n\log(n/k))$ comparisons.