In diborane, boron involves

Diborane consists of 2 boron atoms and 6 hydrogen atoms. Out of these 6 hydrogen atoms 4 hydrogen atoms are attached to only one boron atom remaining two hydrogen atoms are attached to both borons through a special type of bond.
 

Boron has 3 electrons in valence shell; it cannot complete octet directly through covalent bonds. In the excited state of boron, it has 3 unpaired electrons with 4 orbitals which can involve in bonding. In diborane, 2 hydrogen atoms in between two boron atoms are connected to both atoms through a special 3 centered 2 electron bond. In 3 centered 2 electron bonds, 2 electrons are shared in between 3 centers; those are 2 boron atoms and 1 hydrogen atom to satisfy the octet of boron. This type of bond is also considered as regular while calculating hybridization. Since each boron atom has 4 bonds so hybridization will $s{p}^3$ – hybridization.

Hence, option D is correct.