Recently I've been looking at connections between Laver functions on large cardinals and diamonds. While $\diamondsuit$-like principles tend to readily generalize to Laver function-like concepts, I've run into a problem when going the other way.
Let $S\subseteq \omega_1$ be stationary (you can replace $\omega_1$ with a regular $\kappa$ everywhere). Is there a normal uniform $\sigma$-complete filter on $\omega_1$ containing $S$? In particular, is the filter generated by $S$ together with the club filter normal?
After hot_queen resolved my confusion in the comments, I decided to extend the question a bit to try to make it slightly less trivial.
Suppose we have $\omega_1$ many stationary sets $S_\alpha\subseteq \omega_1$ such that any countably many of them have stationary intersection. Is there now a normal uniform $\sigma$-complete filter containing all of the $S_\alpha$? Is the $\sigma$-complete filter generated by the $S_\alpha$ and the club filter normal?
The answer to the first question is yes. For any normal ideal $I$ on any kind of set $Z$ for which normality makes sense, $I \restriction S =_{df} \{ A \subseteq Z : A \cap S \in I \}$ is normal. hot_queen justified this in the comments.
The answer to the second question is no. To see this, we use the fact that for any sequence of stationary sets $\{ A_\alpha : \alpha < \omega_1\}$, the diagonal union of this sequence is the least upper bound of this collection. Using this, we can find a partition of $\omega_1$ into $\{ A_\alpha : \alpha < \omega_1 \}$ many stationary sets such that this forms a maximal antichain in the algebra $\mathcal{P}(\omega_1)/NS$, i.e. every stationary $S$ has stationary intersection with one of them.
Now let $S_\alpha = \bigcup_{\beta \geq \alpha} A_\beta$. Any countable subset of the $S_\alpha$'s has stationary intersection, and they generate a countably complete filter when taken together with the club filter.
However they do not generate a normal filter, because each $A_\alpha$ is in the dual ideal. The diagonal union of the $A_\alpha$'s contains a club because they form a maximal antichain.