If you have $K$ states there are $K$ numbers $\pi_i$ and (for each $n$) $K^2$ numbers $P_{i,j}^n$. The $\pi_i$ are long-term probabilities of being in state $i$ and the $P_{i,j}$ are long-term conditional probabilities of moving from state $j$ to state $i$
There's a close relationship between the sets of numbers: everything is the way it is because it got that way. If there's a stationary distribution and you pick $j$ from that stationary distribution, then $\pi_i=\sum_j P^n_{i,j}\pi_j$ for any $n$ because that's what it means to be a stationary distribution -- if you're in it, you stay in it.
If the chain has a unique stationary distribution and converges to it, then as $n\to\infty$, for any $j$, $P^n_{i,j}\to\pi_i$. Again, that's what it means: that you can start chains anywhere, and you they will eventually end up with close to the stationary distribution. That's what the theorem is saying. But there are assumptions; it's not just true by definition and it's not just two ways of saying the same thing
To see that the assumptions matter, suppose you had a chain with two subsets of states ('red' and 'blue') and it always moved to a state of a different colour at every transition. There would be no vector $\pi_i$ such that $P^n_{i,j}\to\pi_i$ for every $j$, because $P^n_{i,j}$ would always be zero for different-coloured states $(i,j)$ and even $n$, or for same-coloured states $(i,j)$ and odd $n$. That's why the theorem requires that $n$ is chosen to give $P^n_{i,j}>0$, and positive recurrence plus connectedness guarantees this is possible
Conversely, imagine a chain that always transitions to a state of the same colour. In that case, knowing that a red state $i$ is positive recurrent doesn't tell you that a blue state $j$ is positive recurrent because you can't get there from here. Again, the conditions of the theorem are needed for the result.
