While we implement a STACK by a QUEUE, we can make either the PUSH or POP operation $O(1)$ and the other will be $O(n).$ Here, PUSH operation is being favored and so it will be $O(1)$ and POP will be $O(n).$
So, after $k$ PUSH and $l$ POP operations, total operations will be $O(k) + O(ln) = O(ln)$ since $(k << n).$