Consider a discrete Markov source, ${\mathscr{X} \ = \ X_i\}_{i=1}^{\infty}}$ ona finite alphabet set. Let the initial distribution be ${Q}$ and the transition probability for the ${n^{th}}$ step be ${P_n}$. When can we say that ${\mathscr{X}}$ is stationary?
Clearly, the source has to be time invariant and thefore we need ${P_n = P, \forall n}$. For ${\mathscr{X}}$ to be stationary, we need
$\displaystyle f(X_1) = f(X_2) = \ \ldots \ f(X_n)$
etc, where ${f()}$ is the distribution. But ${f(X_n) = QP^n}$. Thus ${Q = QP}$ guarantees that all ${X_n}$‘s have the same distribution. Now, consider, say, ${f(X_1;X_2;X_3)}$ and ${f(X_2;X_3;X_4)}$.
$\displaystyle f(X_1;X_2;X_3) = f(X_1)f(X_2/X_1)f(X_3/X_2).$
$\displaystyle f(X_2;X_3;X_4) = f(X_2)f(X_3/X_2)f(X_4/X_3).$
It is clear that for the two joint distributions to be equal, ${f(X_1) = f(X_2)}$ is enough and therefore ${Q = QP}$ is sufficient.