What is the hybridisation of the central P atom in $\mathrm{PCl_5}$?

P in $\mathrm{PCl_5}$ has five bonded chlorines and no lone pair. That gives five electron domains. The hybridisation needed is $sp^3d$ (one $s$ + three $p$ + one $d$ orbital = five hybrid orbitals). The molecular geometry is trigonal bipyramidal. $sp^3d^2$ would give six domains and an octahedral geometry, which is what you see in $\mathrm{SF_6}$, not $\mathrm{PCl_5}$.