There are two equations here: $\pi P=\pi$ and $\sum_i \pi_i=1$. So you will get eight equations to solve for seven variables. You need to omit one of the equations you get from $\pi P=\pi$ and solve the linear equations.
$$P=\left[\begin{array}{ccccccc}l&m&n&0&0&0&0\\o&p&0&0&0&0&0\\
o&p&0&0&0&0&0\\1&0&0&0&0&0&0\\1&0&0&0&0&0&0\\1&0&0&0&0&0&0\\1&0&0&0&0&0&0\end{array}\right]$$
These are the linear equations you need to solve:
$$l\pi_0+o(\pi_1+\pi_2)+\pi_3+\pi_4+\pi_5+\pi_6=\pi_0\\
m\pi_0+p(\pi_1+\pi_2)=\pi_1\\
n\pi_0=\pi_2\\
\pi_0+\pi_1+\pi_2+\pi_3+\pi_4+\pi_5+\pi_6=1$$
The recurrent classes do not affect each other. Once we get into one recurrent class, we never leave, and the structure of the other recurrent class is irrelevant.
Just take an example, say with 4 states and transition probability matrix $(P_{ij})$ given by:
$$ (P_{ij}) = \left[ \begin{array}{cccc}
0 & 1/2 & 1/2 & 0 \\
0 & 1 & 0 & 0 \\
0 & 0 & 0 & 1 \\
0 & 0 & 1 & 0
\end{array}
\right] $$
State 1 is transient. State 2 forms an aperiodic recurrent class: If we get to state 2 then we always stay there. States 3 and 4 form a periodic recurrent class: If we get to state 3, we bounce around (periodically) between 3 and 4.
So:
-Given we start in state 2: We have a steady state distribution of $\pi=[0,1,0,0]$ (the limiting probabilities converge to this, and the time average fractions of time also converge to this with probability 1).
-Given we start in state 3: The limiting probabilities do not converge (they oscillate depending on even or odd slots), but the time averages converge to $p = [0,0,1/2,1/2]$ with probability 1.
-Given we start in state 1: Then the time averages will certainly converge, but they will not converge to a constant vector with probability 1. Rather, they will converge to a random vector. What they converge to will depend on the outcome of the first transition. If $p=[p_1,p_2,p_3,p_4]$ are the time averages, then given we start in state 1 we get:
$$ p = \left\{ \begin{array}{ll}
[0, 1, 0, 0] &\mbox{ with prob $1/2$} \\
[0, 0, 1/2, 1/2] & \mbox{ with prob $1/2$}
\end{array}
\right. $$
In general, for a finite state discrete time Markov chain with $K$ recurrent classes, each recurrent class $k \in \{1, \ldots, K\}$ has a unique probability distribution $\pi_k$ that satisfies $\pi_k = \pi_k P$ (where $P$ is the transition probability matrix) and such that $\pi_k$ has support only on the states of recurrence class $k$. If we start at a state in recurrence class $k$, then with probability 1 the time averages converge to $\pi_k$. If we start in a transient state, then the time averages will converge to a random vector $p$. We will eventually end up in one of the recurrent states (being the one we first visit). We can define $\theta_k$ as the probability we first visit recurrence class $k$ (defined for each $k \in \{1, \ldots, K\}$). Then $p$ is a random vector with $p = \pi_k$ with probability $\theta_k$, for $k \in \{1, \ldots, K\}$.
Best Answer
Note that two things can go wrong for irregular matrices.
One thing that can go wrong is that the matrix is reducible. What this means for the Markov chain is that there is a non-zero possibility of staying within a proper subsets of the available states. In such a case, we'll end up with a space of steady-state vectors, corresponding to multiple "final distributions".
Another thing that can go wrong is that the matrix is periodic. What this means is that you're caught in cycles of a particular length, so that you can only get back to where you started on steps that are multiples of some period $n$. In such a case, you get most of the same properties that you had before, only now you'll have multiple (complex) eigenvalues of absolute value $1$. There is still, however, a unique steady-state vector.
It turns out that these are the only things that can go wrong, in some combination. For more information on this and everything I've said here, see this wiki page and, in particular, the Frobenius theorem for irreducible matrices.