The key point is that –SO₃H acts as a temporary blocking group.

Step 1: Desulphonation

The starting compound is p-hydroxybenzenesulphonic acid:

On treatment with aqueous bromine, the strongly activating –OH group directs bromination to the ortho and para positions.

But the para position is occupied by –SO₃H.

Step 2: Bromination

Therefore, Br enters the two ortho positions:

The final product is:$$\boxed{\text{2,4,6-Tribromophenol (white precipitate)}}$$

Why does –SO₃H disappear?

Under the aqueous bromination conditions, desulphonation occurs, replacing SO₃H by H. This allows bromination at the para position as well.

So the overall idea is:$$\boxed{\text{Blocking group removed + bromination at all available o/p positions}}$$

Hence 3 Br atoms finally occupy the 2, 4 and 6 positions of phenol.