Let $A \in M_n$ be stochastic and let $\lambda$ be any eigenvalue of A such that |λ| = 1. Why is $A$ power bounded? Let $A \in  M_n$ be stochastic and let $\lambda$ be any eigenvalue of A such that |λ| = 1.
Why is  $A$  power bounded and
conclude that λ is a semisimple eigenvalue of A?
 A: Hint: Consider the row-sum norm, and note that $A^k$ is also stochastic for any integer $k$.
Alternate Hint: $A^k$ is stochastic is for any $k$.  It therefore suffices to note that the set of stochastic matrices is bounded.  That is: for any choice of matrix norm and any fixed $n$, there is a real number $C > 0$ such that $\|A\| < C$ whenever $A$ is a stochastic matrix.
A: We can actually prove something stronger (which I used in another thread):

Proposition. Let $A\in M_n(\mathbb C)$.
(a) If $\rho(A)=\|A\|_\square=1$ for some submultiplicate norm $\|\cdot\|_\square$, then $A$ is power-bounded.
(b) If $\rho(A)=1$, then $A$ is power-bounded if and only if every eigenvalue of $A$ of maximum modulus is semisimple.
(c) $\rho(A)=\|A\|_\square$ for some submultiplicate norm $\|\cdot\|_\square$ if and only if every eigenvalue of $A$ of maximum modulus is semisimple.
Proof. (a) By assumption, $1 = \left(\rho(A)\right)^k=\rho(A^k)\le\|A^k\|_\square\le\|A\|_\square^k=1$. Hence $A$ is power bounded with respect to $\|\cdot\|_\square$. Since all matrix norms (submultiplicative or not) on $M_n(\mathbb C)$ are equivalent, $A$ is power-bounded with respect to every matrix norm.
(b) We shall consider the forward implication only, as the backward implication is obvious. Suppose $A$ is power-bounded. Let $A=PJP^{-1}$ be a Jordan decomposition of $A$. Define a new norm $\|M\|=\|P^{-1}MP\|_F$, where $\|\cdot\|_F$ is the Frobenius norm. As $A$ is power bounded, $\|J^k\|_F=\|A^k\|$ is uniformly bounded. Hence every eigenvalue $\lambda$ of $J$ of unit modulus is semisimple, otherwise $J^k$ would possess an unbounded super-diagonal entry $k\lambda^{k-1}$.
(c) We omit the trivial case $A=0$ and assume that $A$ is nonzero. Suppose $\rho(A)=\|A\|_\square$. Scale $A$ to make $\rho(A)=\|A\|_\square=1$. Hence $A$ is power-bounded by (a) and every eigenvalue of maximum modulus is semisimple by (b).
Conversely, suppose every eigenvalue of $A$ of maximum modulus is semisimple. Let $A=PJP^{-1}$ be a Jordan decomposition of $A$ and let $\Lambda=\operatorname{diag}(1,\epsilon,\epsilon^2,\ldots,\epsilon^{n-1})$ for some $\epsilon>0$. Define $\|X\|_\square = \|\Lambda^{-1}P^{-1}XP\Lambda\|_2$. Then $\|A\|_\square=\|\Lambda^{-1}J\Lambda\|_2$ would become $\rho(A)$ when $\epsilon$ is sufficiently small.

Now, let's go back to your original question. Since your $A$ is stochastic, we have $\rho(A)=\|A\|_\infty=1$, where $\|\cdot\|_\infty$ is the maximum row sum norm (i.e. the submultiplicative matrix norm induced by the vector $\ell_\infty$ norm). Hence the result follows from parts (a) and (c) in the proposition above.
