Alcoholic KOH performs $$\alpha $$ - $$\beta $$ elimination with halides.

The carbon which is attached to Br atom is called $$\alpha $$ carbon.

Here, two $$\alpha $$ carbon presents. Let's call them $$\alpha $$₁ and $$\alpha $$₂.

In case of $$\alpha $$₁ carbon, H elimination can take place either from $$\beta $$₁ or $$\beta $$₂ carbon. But as we know always more stable product is formed from a reaction. Here if H is removed from $$\beta $$₁ carbon then a $$\pi $$ bond is created which will perticipate in resonance with benzene ring and product will be more stable. But if H is removed from $$\beta $$₂ carbon then the created $$\pi $$ bond will not perticipate in resonance with benzene ring so product will be less stable.

In case of $$\alpha $$₂ carbon, H elimination can take place either from $$\beta $$₂ or $$\beta $$₃ carbon. But as we know always more stable product is formed from a reaction. Here if H is removed from $$\beta $$₂ carbon then a $$\pi $$ bond is created which will perticipate in resonance with the $$\pi $$ bond associated with the $$\alpha $$₁ carbon so product will be more stable. But if H is removed from $$\beta $$₃ carbon then the created $$\pi $$ bond will not perticipate in any resonance so product will be less stable.