Since you are already given an stationary distribution, you should only check it. For the definition of the stationary distribution you should have $$ \pi(A) = \int\limits_{\mathbb R}P(y,A)\pi(dy) $$ for any measurable set $A$. So you know that $$ P(y,A) = \int\limits_{A}\frac1{\sigma\sqrt{2\pi}}\mathrm e^{-\frac{(z-y/2)^2}{2\sigma^2}}\,dz. $$ Where $\sigma^2 = 3/4$.
So you should show that $$ \int\limits_{A}\frac1{\sqrt{2\pi}}\mathrm e^{-\frac{y^2}{2}}\,dy = \int\limits_{\mathbb R}\frac1{\sqrt{2\pi}}\mathrm e^{-\frac{y^2}{2}}\,dy\int\limits_{A}\frac1{\sigma\sqrt{2\pi}}\mathrm e^{-\frac{(z-y/2)^2}{2\sigma^2}}\,dz $$
Let as work a bit with the right-hand side: $$ \frac1{2\sigma\pi}\int\limits_{\mathbb R}\mathrm \,dy\int\limits_{A}\mathrm e^{-\frac{(z-y/2)^2}{2\sigma^2}+y^2/2}\,dz = \frac1{2\sigma\pi}\int\limits_{\mathbb R}\mathrm \,dy\int\limits_{A}\exp\left\{-\frac{(z-y/2)^2+\sigma^2y^2}{2\sigma^2}\right\}\,dz $$ $$ = \frac1{2\sigma\pi}\int\limits_{\mathbb R}\mathrm \,dy\int\limits_{A}\exp\left\{-\frac{z^2-zy+y^2}{2\sigma^2}\right\}\,dz= \frac1{2\sigma\pi}\int\limits_{\mathbb R}\mathrm \,dy\int\limits_{A}\exp\left\{-\frac{(y-z/2)^2+\sigma^2z^2}{2\sigma^2}\right\}\,dz $$ and here we change the integrals (do you know why is it allowed?) $$ = \frac{1}{\sqrt{2\pi}} \int\limits_{A}\mathrm e^{-z^2/2}\,dz\int\limits_{\mathbb R}\frac{1}{\sigma\sqrt{2\pi}}\mathrm e^{-\frac{(y-z/2)^2}{2\sigma^2}}\,dy = \frac{1}{\sqrt{2\pi}} \int\limits_{A}\mathrm e^{-z^2/2}\,dz = \pi(A) $$ since the internal integral is always $1$ due to the fact it's an integration ofPDF on the whole state space.
About the irreducibility: we should show that if $\lambda(A)>0$ then the return probability $L(x,A)>0$ for all $x\in \mathbb R$. The simple estimation $L(x,A)\geq P(x,A)>0$ for all $A:\lambda(A)>0$ will lead you to the desired answer.