Let $P$ be a partition of a group with $AB \subseteq C$. Why is $1 \in P_n$? $P_n$ is the equivalence class of $n \in N$ and $1 \in N=P_1$.

Question

Let P be a partition of a group G with the property that for any pair of elements A, B of the partition, the product set AB is contained entirely within another element C of the partition. Let N be the element of P that contains 1. Prove that N is a normal subgroup of G and that P is the set of its cosets.

This exercise was asked about in the following questions: partition of a group to have normal subgroup partition of a group and cosets Proving a partition is set of cosets Artin 2.10.3 help understanding why $AN=NA=A$ implies $N$ is normal

My question is about Brian Bi's proof linked here, where it is claimed that $1 \in P_n$.

The following is a screenshot of the proof (Kiefer Sutherland's voice):

Please explain the $1 \in P_n$. This is the only part I don't understand.

Answer 1

$n$ belongs to some element $P_n$ of the partition $P$.

$n$ belongs to $N$.

We are given that $N$ is not just any subset of $G$: $N$ is also an element of $P$.

$\therefore, N \cap P_n \ne \emptyset \implies N = P_n$

$1 \in P_1 = N \implies \therefore, 1 \in P_n$.