= Solution
For $a<0$, set $v_\infty=-\sigma^2/(2a)>0$. The candidate <invariant probability law of a Markov process> is
$$
\boxed{\pi=\mathcal N\left(0,-\frac{\sigma^2}{2a}\right),\qquad
\pi(dx)=\sqrt{\frac{-a}{\pi\sigma^2}}\exp\left(\frac{ax^2}{\sigma^2}\right)\,dx.}
$$
To prove invariance rather than merely guess a limiting <probability distribution>, initialize $X_t\sim\pi$ independently of future <Brownian increments>. Integrating the <conditional characteristic function> from part (i) against this centered <normal distribution> gives
$$
\begin{aligned}
\mathbb E[e^{iuX_{t+\Delta}}]
&=e^{-u^2v_\Delta/2}\,\mathbb E[e^{iu e^{a\Delta}X_t}]\\
&=\exp\left[-\frac12u^2\left(v_\Delta+e^{2a\Delta}v_\infty\right)\right]
=e^{-u^2v_\infty/2},
\end{aligned}
$$
because $v_\Delta=v_\infty(1-e^{2a\Delta})$. The <uniqueness theorem for characteristic functions> implies $X_{t+\Delta}\sim\pi$ for every $\Delta\geq0$. This proves $\pi P_\Delta=\pi$, as required. The <time-homogeneous Markov property> then also makes the process with this <Ornstein-Uhlenbeck stationary initialization> a <strictly stationary process>.
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