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
The solution is wrong. Having been told that you start at $3$ and given this chain, the steady state distribution is $\frac{2}{5}(4/5,1/5,0,0) + \frac{3}{5}(0,0,0,1)=(8/25,2/25,0,3/5)$. This changes if you change the initial distribution (because there are two independent invariant distributions as a result of there being two recurrent classes), but that doesn't mean it simply doesn't exist.
The distribution after $10^{10}$ steps deviates from this by a tiny amount; in particular $p_3$ at this point is $2^{-10^{10}}$. The other probabilities will deviate by similar amounts. For this to work it was important that there is a self-edge from $1$ to $1$, so that the dynamics confined to $\{ 1,2 \}$ are aperiodic.