Example of the $Ω$−partite Consider the following definition from section 2.2 of this research paper

We write $\mathcal{P}()$ for the power set of the set $$. An $-$partition
$Ω=\{_1,\cdots,_\}$ of a set $$ is a disjoint family of $$ nonempty
subsets of $$ with $=_1∪ \cdots ∪ _$. Let $Λ⊆\mathcal{P}()$ be a family of
subsets of $$. For a permutation $$ on $$, we define
$(Λ)=\{():∈Λ\}⊆\mathcal{P}()$. A family of subsets $Λ⊆\mathcal{P}()$ is said to
be $Ω-$partite if $(Λ)=Λ$ for every permutation $$ such that
$(_)=_$ for every $_∈Ω$. We say that $Λ$ is $-$partite if
it is $Ω-$partite for some $-$partition $Ω$. These concepts can be
applied to access structures, which are actually families of subsets.

My doubt lies in $(Λ)=\{():∈Λ\}⊆\mathcal{P}()$ because the paper does not define the permutation function on a set, but used it. I didn't came through any standard defifnition of permutation function on set which gives another set which is different from the input set because the order of elements in a set is immaterial.
What I am clear from the context is
$$\sigma: P \rightarrow P$$
$$ \sigma(A \in \mathcal{P}(P)) = B \in \mathcal{P}(P)$$
but I don't know what exactly $B$ is.
What is the definition of a permutation function over a set?
In addition, can you give some example of the above $Ω-$partite ?
A permutation on a set is a bijection from the set to itself; thus $\sigma$ is a (bijective) map from $P$ to itself, not from $p(P)$ to itself (though it does induce a map from $p(P)$ to itself in a natural manner).

Also, if $f$ is a map from $A$ to $B$ and $X$ is a subset of $A$, then $f(X) = \{f(x), x \in X\}$. This is completely standard notation.

For a collection $\Lambda$ of subsets of $P$, $\sigma(\Lambda) = \{\sigma(A), A \in \Lambda\}$ is the collection of subsets of $P$ which is obtained by applying $\sigma$ to all the elements of $\Lambda$.

Example: $P = \{1,2,\dots,8\}$, $\Omega = \{\{1,2\},\dots,\{7,8\}\}$. Then $\Lambda = \{\{1,2,3,4\},\{5,6,7,8\}\}$ is $\Omega$-partite since clearly every permutation of $P$ that fixes each $P_i$ also fixes each element of $\Lambda$, so in fact we even have $\sigma(A) = A$ for all $A$. On the other hand, $\Lambda' = \{\{1,2,3\},\{4,5,6\},\{7,8\}\}$ is not; for instance $\sigma = (34)$ yields $\sigma(\Lambda') = \{\{1,2,4\},\{3,5,6\},\{7,8\}\} \ne \Lambda'$.