If the friends take $m$ photos, there are exactly $m(n-1)$ appearances in total. Everyone needs at least $k$ appearances, so $m(n-1) \ge nk$. Therefore, at least $\lceil nk/(n-1)\rceil$ photos are necessary.

Taking turns

Let $m=\lceil nk/(n-1)\rceil$, and let friends take turns being the photographer in the order $1,2,\ldots,n,1,2,\ldots$.

Each friend takes at most $\lceil m/n\rceil$ photos. From $m(n-1)\ge nk$, we obtain $m/n\le m-k$. Since $m-k$ is an integer, $\lceil m/n\rceil\le m-k$. Thus, every friend appears in at least $m-\lceil m/n\rceil\ge k$ photos.

The lower bound is achievable, so $\mathrm{answer}=\lceil nk/(n-1)\rceil$. Using integer division, compute it as $(nk+n-2)/(n-1)$.

The time complexity is $O(1)$, and the extra space complexity is $O(1)$.