Boron has three valence electrons so it is supposed to make 3 bonds in a molecule with hybridisation sp2 as only s and two p orbitals are used in hybridisation because last p orbital is vacant.
But case of diborane is different. In diborane bonding of B atom with H and other B is different. Is has two three-center 2-electron bonds, each one consisting of the two boron atoms and a hydrogen atom sharing two electrons.
The molecule has a total of 12 valence electrons, 3 from each boron atom and 1 from each hydrogen atom. In this molecule, each of the two boron atoms will be bonded to two hydrogen atoms via covalent bonds.These bonds will use up 8 of the 12 valence electrons.
The other 4 valence electrons will be shared by the remaining two hydrogen atoms and by the boron atoms, each atom contributing one orbital to the formation of the 3-center 2-electron bond. Such bond is called banana bond.
Thus 4 orbitals of each boron atom will be hybridised to accommodate are the electrons (electrons of 2 B-H covalent bonds and shared electrons of 2 banana bonds). Thus hybridisation of boron in diborane is sp3.