Answer
It is not a stationary state.
Work Step by Step
$$|\Psi(x)|^2=\Psi^*\Psi$$
$$\Psi^* = \psi^*\sin \omega t$$
$$|\Psi(x)|^2=|\psi|^2\sin^2 \omega t$$
$|\Psi(x)|^2=\Psi^*\Psi$ is not time-independent, so $\Psi$ is not the wave function for a stationary state.