Let $p+q+r=1, p, q, r > 0.$ Let $M = \begin{bmatrix}p&q&r\\ r&p&q\\q&r&p\end{bmatrix} = PDP^{-1}$ where $D = \text{diag}(1, \alpha, \beta)$ (this is shown using Perron-Frobenius, and I suspect there's no clean proof without the theorem) where $|\alpha|,|\beta|<1.$
Then $\lim\limits_{n \to \infty} M^n = P\text{diag}(1,0,0)P^{-1}$ is claimed to be a matrix where all of the entries are equal. This is quite a leap of logic. Here is what I suspect was left out:
Let the first row of $P^{-1}$ be $(a,b,c).$ We may pick $P$ so that the first column of $P$ is $(1,1,1),$ which makes the matrix $\begin{bmatrix}a&b&c\\ a&b&c\\a&b&c\end{bmatrix}.$ Thus, we need to prove $a=b=c.$ But how? From $I=P^{-1}P,$ we have $a+b+c=1,$ which is not enough information.
Edit: Let $P = \begin{bmatrix} v_1 & v_2 & v_3 \end{bmatrix}, N = P\text{diag}(1,0,0)P^{-1}.$ Then $v_2, v_3$ are linearly independent eigenvectors in the null space of $N,$ so $\dim \text{Nul} \, N \ge 2 \Rightarrow \dim \ker N \le 1,$ which implies the result. This is too much to simply leave out, so I think whoever wrote the proof did a shoddy job near the end.
Explanation : eigenvalue $1$ has
$$\begin{bmatrix}1\\1\\1\end{bmatrix}$$
as an associated eigenvector. Indeed
$$\begin{bmatrix}p&q&r\\ r&p&q\\q&r&p\end{bmatrix}\begin{bmatrix}1\\1\\1\end{bmatrix}=\underbrace{(p+q+r)}_{ = 1}\begin{bmatrix}1\\1\\1\end{bmatrix}$$
That's all...