A transition matrix $P$ is said to be doubly stochastic if the sum over each column equals one; that is,
$$\sum_{i}P_{ij}=1, \quad \text{ for all } j$$
If such a chain is irreducible and consists of $M + 1$ states $0, 1, \dots , M$, show that the long-run proportions are given by $$\pi_j = \frac{1}{M+1}, \quad\quad j = 0, 1, \dots , M$$
Here I think we use the fact that $P$ is doubly stochastic. Also, I think I have to show is aperiodic but I honestly don't know how?
The result holds irrespective of whether the chain is periodic, though being irreducible of course does matter quite a bit. For an nice periodic example to apply this to, consider the $\text{(m + 1) x (m + 1) }$ cyclic shift operator (a permutation matrix)).
Here are two proofs:
1.) The conclusion is an immediate consequence of the Elementary Renewal Theorem and the fact that $\pi_{j} = \frac{1}{E[X]}$ (where $X$ is a positive integer valued random variable that takes on values associated with the first hitting time of visiting state $j$ given a start in state $j$. Supposing that $\pi_{j}$ exists, the fact that $\pi_{j} = \frac{1}{E[X]}$ is itself an immediate consequence of applying the elementary renewal theorem to a delayed renewal process -- i.e. one that is started in steady state.)
2.) For an algebraic approach, consider computing the Cesaro Sum
$\frac{1}{n}S_n = \frac{1}{n}\big(I + P + P^2 +... +P^{n-1}\big)$ with the goal of showing $\frac{1}{n}S_n \to E = \mathbf {1\mathbf \pi}^T$
This should remind you of telescoping a finite geometric series, though there are some singularity issues to deal with. (This telescoping argument is an exercise in the Markov Chains chapter of Grinstead and Snell's free intro probability book as I recall.)
To telescope:
first notice commutativity: $P^kE = EP^k$ then
$\frac{1}{n}S_n\big(I-P+E\big) $
$= \frac{1}{n}\big(I + P + P^2 +... +P^{n-1}\big)\big(I-P+E\big) $
$= \frac{1}{n}\big(I-P^n + n E\big)$
$= \frac{1}{n}I + \frac{1}{n}P + E$
it should be clear that
$\frac{1}{n}S_n\big(I-P+E\big) \to E$
it remains to verify that $\big(I-P+E\big)^{-1}$ exists and that $\big(I-P+E\big)^{-1}E = E$. The latter is immediate because $\big(I-P+E\big)E = E - E + E = E$
As for showing that $\big(I-P+E\big)^{-1}$ exists, this depends on what machinery you know. The result is pretty immediate if you know much about Perron Frobenius theory or Jordan Canonical Forms. In your particular problem since your matrix is doubly stochastic, you could even apply Schur Triangularization, selecting the ones vector as the left vector $\mathbf q_0$ in $Q$ where $P=QTQ^*$ and observe that the triangularized matrix has no zeros on the diagonal and hence is invertible. There are still other ways to approach this e.g. since $\dim \ker \big(I-P\big)=1$, directly show that $\big(I-P + E\big)$ has a trivial kernel.