Support there is a Array A with size n and let n be a multiple of k. If we divide the A into sub arrays each with elements k and sort them individually, it would require k*log(k) time. Total initial time =(n/k)klog(k)=nlog(k). And then merging the sorted sub arrays, let say using a heap of size k will cost atmost nlog(k) time. So total time = 2*nlog(k)+c which is O(nlog(k)). But we know that lower bound for sorting is n*log(n). What is wrong in my thinking?
The problem comes from merging the arrays.
What I think you're thinking when using a heap is to keep track of the smallest elements from each of the $k$-element sorted arrays, removing the smallest element $s$ them from the heap, putting $s$ in the output, and adding a new element onto the heap from the array which contained $s$.
The problem with this is that there are $n/k$ arrays, so your heap will actually contain $n/k$ elements.
As an extreme case, let $k=1$. Then your algorithm would consist of sorting each 1-element array (which takes no time) and then merging them using heapsort.