Choose the correct set of reagents for the following conversion
trans Ph - CH = CH - CH$$_3$$ $$\rightarrow$$ cis Ph - CH = CH - CH$$_3$$

We need to convert trans-PhCH=CHCH$$_3$$ to cis-PhCH=CHCH$$_3$$.

Trans to cis conversion cannot be done directly. The approach is:

1. Convert the double bond to a triple bond (via addition-elimination)

2. Then selectively reduce the triple bond to give the cis alkene

1. Br$$_2$$: Anti addition to the trans alkene gives a vicinal dibromide

2. Alc. KOH: Double dehydrohalogenation (two eliminations) gives the alkyne PhC$$\equiv$$CCH$$_3$$

3. NaNH$$_2$$: Strong base for the second elimination (if needed) to ensure complete conversion to alkyne

4. H$$_2$$/Lindlar catalyst: Selective syn-hydrogenation of the alkyne gives the cis alkene

The key is that Lindlar's catalyst (Pd/CaCO$$_3$$/quinoline) gives cis (syn) addition, while Na in liquid NH$$_3$$ (Birch conditions) would give the trans product.

Option B: Br$$_2$$, alc KOH, NaNH$$_2$$, H$$_2$$ Lindlar Catalyst - this correctly uses Lindlar's catalyst for cis selectivity.

The correct answer is Option B: Br$$_2$$, alc KOH, NaNH$$_2$$, H$$_2$$ Lindlar Catalyst.