we know that both cases have been covered in duplicates keys and distinct keys
Assume root is at first level and height of root is $0$.
the first min can be found at most on the first level, 2$^{nd}$ min can be found at most on the second level, 3$^{rd}$min can be found at most on the third level...... so pth min element can be found at pth level.similarly, on deleting the pth min, we got a pth minimum $( p<n )$ which we want.
So I need not search in full tree.
on deleting root, we got the first min.then again adjust heap then again deleting root, we got the second min and then adjust heap.
and so on we do this by deleting pth root element of the tree we got a pth minimum value.
So I have to apply this operation p times and each operation will take $0$$\left ( \log p \right )$Otime.
So overall time complexity will be : $0$$\left ( p\log p \right )$