The correct shape and hybridisation for $\text{XeF}_4$ are:

In $\text{XeF}_4$, the central Xenon (Xe) atom has 8 valence electrons. It forms 4 single bonds with 4 Fluorine (F) atoms using 4 of its valence electrons. The remaining 4 valence electrons form 2 lone pairs. The steric number (total electron domains) = Number of bond pairs (4) + Number of lone pairs (2) = 6. A steric number of 6 corresponds to $sp^3d^2$ hybridisation. According to VSEPR theory, the 6 electron domains arrange themselves in an octahedral geometry to minimize repulsion. To minimize the strong lone pair-lone pair repulsions, the two lone pairs occupy opposite axial positions. This leaves the 4 Fluorine atoms in the equatorial positions, resulting in a square planar molecular shape.