In order to understand recurrence, transience, non-return state, and absorbing state, we don't require the actual transition probabilities of a Markov Chain, if the states of the chain can be accommodated in state transition diagram. For example, to understand the nature of the states of above Markov Chain, the given transition matrix can be equivalently be represented as
\begin{equation*}
P = \left(\begin{array}{ccc}
* & * & *\\
0 & * & *\\
0 & 0 & *\\
\end{array}\right)
\end{equation*}
where a * stands for positive probability for that transition.
Now, draw the state transition diagram of the Markov Chain.
There are 3 communicating classes, here: {1}, {2} and {3}. Now identify which of these classes are closed communicating classes and non-closed communicating classes.
Consider class {1}. State 1 communicates with itself. However, an escape is possible to state 2 or state 3. Hence, it is a non-closed communicating class. States in a non-closed communicating classes become transient states.
Class {2} can be interpreted in a similar manner.
State 3 communicates with itself and all the edges are into the state 3. Hence, state 3 itself forms a closed-communicating class. States in a closed communicating classes become recurrent states. As there is only one state in this communicating class, the state is called an absorbing state. In a finite Markov Chain, there must be at least one recurrent state. As all the states do not belong to a single communicating class, the given chain is not irreducible.
This answer is based on these lecture notes.
Let $P$ be the transition matrix, with $(P)_{i,j} = p_{i,j}$. A chain is irreducible if for all $i$,$j$, there is some $t$ such that $(P^t)_{i,j} > 0$. Since $P$ is tridiagonal and all entries are positive, it is irreducible.
Now from the notes, we have the proposition (Excercise 2.9): "Suppose $P$ is irreducible and contains at least one self-loop (i.e., $(P)_{i,i} > 0$ for some $i$). Then $P$ is aperiodic."
Clearly $P$ has self-loops, since $p_{i,j} \in (0,\frac{1}{4})$ if $i\neq j$ and $\lvert i-j\rvert \le 2$, and from the definition:
$$p_{i,i} = 1 - p_{i,i-1} - p_{i, i-2} - p_{i,i+1} - p_{i,i+2} > 1 - \frac{1}{4} - \frac{1}{4} - \frac{1}{4} - \frac{1}{4} = 0$$
Putting this together with the proposition gives your aperiodicity result.
Best Answer
Definition Let $p_{ii}^{(n)}$ denote the probability of returning to state $i$ at step $n$ and let $t\in\{2, 3\dots\}$. State $i$ is said to be periodic with period $t$ iff
If we can not find a $t$ such that this holds, the state is said to be aperiodic.
Solution In your case it would be useful to draw a transition diagram of the matrix. You can see that if the chain starts in $c$ then returns to $c$ are possible at steps $2, 3, 4, 5, \dots$. As we can not find a $t$ such that the definition holds, $c$ is an aperiodic state.