The stem carries a drawn structure; the four options are text and are reproduced in full above.

Two descriptors are wanted — R/S at the stereocentre and E/Z at the double bond — and both come from the same Cahn-Ingold-Prelog priority rules.

Assigning R or S. Rank the four groups at the stereocentre by atomic number at the first point of difference, put the lowest priority group away from you, and read 1→2→3: clockwise is R, anticlockwise is S. Here bromine (Z = 35) clearly outranks every carbon substituent, so it takes priority 1.

Assigning E or Z. Rank the two groups on each carbon of the double bond separately. If the two higher-priority groups are on the same side the alkene is Z; on opposite sides, E.

The subtlety this question is really testing is isotopes. The molecule contains both C₆H₅ (ordinary phenyl) and C₆D₅ (perdeuterated phenyl). These are identical in atomic number, so the usual first test cannot separate them. CIP resolves such ties with a later rule: where atomic numbers are equal, the higher mass number takes priority. Deuterium (mass 2) therefore outranks protium (mass 1), and C₆D₅ ranks above C₆H₅.

That single decision is what makes the molecule stereochemically defined at all — without isotopic substitution the two phenyl groups would be identical and there would be no stereocentre to describe. The ester group CO₂Et outranks both phenyls, because duplicated oxygen atoms at the first carbon beat the carbons of an aromatic ring.

Per the official final answer key the answer is option (A), S and E.