Consider a Markov chain with state space $\{0,1,2,3,4,5\}$ and transition probability matrix $$P= \begin{pmatrix} 0&1&0&0&0&0\\ 1/5&0&4/5&0&0&0\\ 0&2/5&0&3/5&0&0\\ 0&0&3/5&0&2/5&0\\ 0&0&0&4/5&0&1/5\\ 0&0&0&0&1&0 \end{pmatrix} $$ One can check directly that $P$ has a stationary distribution $\pi$ with $$ \pi=\bigg(\frac{1}{32},\frac{5}{32},\frac{10}{32},\frac{10}{32},\frac{5}{32},\frac{1}{32}\bigg). $$ A theorem says that the stationary distribution for a regular matrix is also its limiting distribution. Unfortunately, $P$ is not a regular matrix (due to the zero diagonal), and thus the theorem is not applicable. But I don't know if the "regular" condition could be relaxed or not. Here is my question:
Is $\pi$ also the limiting distribution of $P$?
The usual theorem for convergence to the stationary distribution for a finite state space uses irreducibility and aperiodicity. Irreducibility in your case is clear: the accessibility graph is just the undirected straight line graph which is strongly connected. However this chain is actually periodic. For example, all paths from state 1 to state 1 have a length which is a multiple of 2. This is reflected in an eigenvalue of -1 in the transition probability matrix. Accordingly you have convergence to the stationary distribution if and only if your initial distribution is orthogonal to the right eigenvector with this eigenvalue, which turns out to be $\begin{bmatrix} 1 \\ -1 \\ 1 \\ -1 \\ 1 \\ -1 \end{bmatrix}$. So if $p_1+p_3+p_5=p_2+p_4+p_6$ then you have convergence to the stationary distribution, otherwise you don't.
Note that the theorem you stated is a special case of this one, because having all nonzero diagonal entries gives aperiodicity for free.
Also note that you do have Cesaro convergence to the stationary distribution, i.e. $\lim_{n \to \infty} \frac{1}{n} \sum_{k=1}^n p^{(k)} = \pi$.