The compound shown is one enantiomer of 1-chloro-1-phenylethane. The compound shown is best described as

The benzylic carbon carries $\mathrm{H}$, $\mathrm{Cl}$, $\mathrm{CH}{}_3$ and $\mathrm{C}{}_6\mathrm{H}{}_5$, four different groups, so it is a stereocentre. Planarity of the phenyl ring does not create a mirror plane for the whole molecule because the other three groups are different. The compound is chiral and exists as a pair of enantiomers; there is no second stereocentre, so no meso form, and no $\mathrm{C}=\mathrm{C}$ with two different groups on each end, so no geometrical isomerism.