A machine has three critical parts (1,2,3) but can function as long as two of these parts are functional. When two are broken, they are replaced and the machine is functional the next day. The state space is the parts that are broken {0,1,2,3,12,13,23}. Parts 1,2, and 3 fail with probabilities 0.01, 0.02, and 0.04, respectively, but no two parts fail on the same day.
What is the number of parts of types 1,2, and 3 we will use if the machine operates for 1800 days.
So I understand that the transition probability is given by (where $p(i,j)$ is the probability of going from state $i$ and day $n$ to state $j$ on day $n+1$):
$$\begin{matrix} \quad& 0\quad & 1 \quad& 2\quad&3\quad&12\quad &13\quad&23\\ \\ 0\quad & 0.93\quad & 0.01\quad &0.02\quad&0.04\quad&0\quad&0\quad&0 \\ \\ 1\quad & 0\quad & 0.94\quad&0\quad&0\quad&0.02\quad&0.04\quad&0\\ \\ 2\quad&0\quad&0\quad&0.95\quad&0\quad&0.01\quad&0\quad&0.04\\ \\ 3\quad&0\quad&0\quad&0\quad&0.97\quad&0\quad&0.01\quad&0.02\\ \\ 12\quad&1\quad&0\quad&0\quad&0\quad&0\quad&0\quad&0\\ \\ 13\quad&1\quad&0\quad&0\quad&0\quad&0\quad&0\quad&0\\ \\ 23\quad&1\quad & 0 \quad& 0\quad&0\quad&0\quad &0\quad&0 \end{matrix}$$
So, I know by the convergence theorem, $p^n(x,y)\rightarrow \pi (y)$ as $n\rightarrow \infty$ if $p$ is irreducible and aperiodic. However, for $p$ to be aperiodic, all states $x$ must have $p(x,x)>0$, which is not the case for states 12,13, and 23. So I am lost at what to do.
You have misunderstood. For large enough values of $n$, you will find that $p^n(x,x)>0$ for states 12, 13 and 23.
Look at this: https://www.probabilitycourse.com/chapter11/11_2_4_classification_of_states.php