The compound shown is hex-4-ene-2,3-diol drawn without stereochemistry. Counting geometrical and optical isomers together, the total number of stereoisomers is

C2 carries $\mathrm{H}$, $\mathrm{OH}$, $\mathrm{CH}{}_3$ and C3; C3 carries $\mathrm{H}$, $\mathrm{OH}$, C2 and the propenyl group. Both are stereocentres. The $\mathrm{C}{}_4=\mathrm{C}{}_5$ bond has $\mathrm{H}$ and C3 on C4 and $\mathrm{H}$ and $\mathrm{CH}{}_3$ on C5, so it is $E$ or $Z$. The two ends of the molecule are different, so there is no meso form and no coincidence: total $2\times 2\times 2=8$ stereoisomers.