I RESOLVED THIS QUESTION FOR MYSELF, THANKS FOR VIEWING, SORRY
We have this notation:
$$
\mathcal A \to \mathcal B_k^{n}
$$
Which means:
\begin{align*}
&(\forall A \in \mathcal A)(\forall f)\left(f:[A]^{n}\to \{1,2,\dots,k\}\Longrightarrow \\(\exists B\subseteq A)(\exists i \in \{1,\dots,k\})(B\in \mathcal B \land f \text{ has value $i$ on } [B]^{n})\right)
\end{align*}
To me this is saying for every set from the family $\mathcal A$, for every function from the set of all partitions of $A$ with length $n$ into $\{1,2,\dots, k\}$ there is a subset of $A$, namely $B$ such that $B$ has constant value $i$ on every partition of length $n$ of it and $B$ is a member of $\mathcal B$. Am I interpreting this notation correctly? Is this saying that there is a definitive member $B \in \mathcal B$ that can be colored $1$ or $2$ or $3$ or $4$$\dots$ or $k$ given that we partition it into $n$ many pieces and it is some subset of $A\in \mathcal A$?
I'm assuming that:
$[A]^n$ mean the set of all partitions of $A$ with cardinality $n$. Am I correct in this interpretation?
I RESOLVED THIS QUESTION FOR MYSELF, THANKS FOR VIEWING, SORRY
WRONG $[X]^{n}$ in this context means the set of all size $n$ subsets of $X$.
E.G. $$ 6\to3_2^{2} $$
means for every set of cardinality $6$, for all functions from $f:[6]^{2}\to \{1,2\}$ there exists a $B\subset 6$, such that $[B]^2$ has constant color and $B\subset 3$.