Solution

In hybridization, the number of hybrid orbitals is equal to the number of atomic orbitals that are hybridized.
Let's take an example:- $$CH_4$$
Carbon electronic configuration is $$2S^2$$2P^2$$ in ground state, in excited state one electron of $$s$$ orbital move into $$p$$ orbital and thus we get $$4$$ unpaired electrons in atomic orbital which will take part in bonding,
The hybridisation of carbon in $$CH_4$$ is $$sp^3$$ , means $$4$$ hybrid orbitals. (one $$s$$ orbital and three $$p$$ orbitals)
So the correct option is [B]