What happens to the dual filter when $I=P(S)$ for an ideal of a set $S ?$ $S$ is a set and $I$ is an ideal of $S$ that is not proper that is $I=P(S).$
We define the Dual Filter of $I$ as $$F(I)=\{A\subset S:S\backslash A\in I\}.$$
Now since $I=P(S)$ implies that $S\in I$ then $S\backslash S=\Phi\in F(I)?$ But that can't be because a Filter cannot have $\Phi$  in it.
Then in this paper $I^K$ Convergence by Martin Sleziak & Martin Macaj it's said that: 
If $I$ is not proper, that is, if $I = P(S)$, then every function $f : S → X$ converges to each
point of $X$.
 Wherer $I$-convergence defined by 
Definition $2.1.$ Let $I$ be an ideal on a set $S$ and $X$ be a topological space. A function $f : S → X$ is said to be $I$-convergent to $x ∈ X$ if
$$f^{−1}(U) = \{s ∈ S; f(s) ∈ U\} ∈ F(I)$$
holds for every neighborhood $U$ of the point $x$.
Since $I=P(S)$, every subset of $S$ is in $I$, hence so is the set $A=\{s\in S:f(s)\in U\}$ for some nbd $U$ of $x.$ But since $S\backslash A\in I$ too so we can say $A\in F(I)$ and the criterion for continuity is satisfied, is it?
But what happens with the $\Phi \in F(I) ?$
 A: I assume you're using $\Phi$ as a notation for the empty set. It's better to use $\emptyset$ (\emptyset) or $\varnothing$ (\varnothing). 
According to the usual definition, a filter is allowed to contain $\emptyset$. Just like for ideals, we say that a filter $F$ is proper if $\emptyset\notin F$. Then $I$ is a proper ideal if and only if $F(I)$ is a proper filter. 
Some sources include the condition $\emptyset\notin F$ in the definition of filter on $S$, but these sources should also include the condition $S\notin I$ in the definition of ideal on $S$!
I don't understand your question "What happens with the $\emptyset$ in $F(I)$?". The presence of the empty set in $F(I)$ doesn't cause any problems. As you note, if $I = \mathcal{P}(S)$, then $F(I) = \mathcal{P}(S)$, and the condition in Definition 2.1 is automatically satisfied, i.e. every point in $X$ is an $I$-limit point of every function $f\colon S\to X$. This includes points $x\in X$ which have a neighborhood $U$ which does not contain any points in the image of $f$. For such a neighborhood $U$, we'll have $f^{-1}(U) = \emptyset\in F(I)$. 
