= Solution
A <stationary distribution> is a nonnegative row vector $\pi$ with $\pi P=\pi$ and $\sum_i\pi_i=1$. The balance equations give
$$
\pi_1=\tfrac12\pi_1+\tfrac12\pi_4,\quad
\pi_2=\tfrac12\pi_3+\tfrac14\pi_4,\quad
\pi_3=\pi_2+\tfrac14\pi_4,\quad
\pi_4=\tfrac12\pi_1+\tfrac12\pi_3.
$$
The first yields $\pi_1=\pi_4$. Substitution into the middle pair yields $\pi_2=3\pi_4/4$ and $\pi_3=\pi_4$; the fourth equation is then satisfied. Normalization gives $15\pi_4/4=1$, hence
$$
\boxed{\pi=(4,3,4,4)/15}.
$$
All states communicate, so this is the unique <stationary distribution>. The self-loop at state 1 additionally makes this finite <Markov chain> aperiodic, although aperiodicity is not required for the balance calculation.
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