Recall the meaning of $sp^3d^2$ hybridization

$sp^3d^2$ hybridization corresponds to six electron pairs / six hybrid orbitals around the central atom, which is typically associated with an octahedral electron-pair geometry.

Check each option

A: $BrF_5$

• Central atom: $Br$
• Valence electrons on $Br = 7$
• It forms $5$ $Br-F$ bonds and has $1$ lone pair.
• So, steric number $= 5 + 1 = 6$.
• Hence, it uses $sp^3d^2$ hybridization.

Therefore, A displays $sp^3d^2$ hybridization.

B: $SF_6$

• Central atom: $S$
• It forms $6$ $S-F$ bonds and has no lone pair.
• Steric number $= 6$.
• Geometry is octahedral.
• Hence, $sp^3d^2$ hybridization.

Therefore, B displays $sp^3d^2$ hybridization.

C: $[CrF_6]^{3-}$

• This is a coordination complex.
• Oxidation state of $Cr$: $x + 6(-1) = -3 \Rightarrow x = +3$
• So, $Cr^{3+}$ has configuration: $Cr: [Ar]3d^54s^1$ $Cr^{3+}: [Ar]3d^3$
• With six ligands, the complex is octahedral.
• In traditional hybridization language, octahedral complexes are described as $d^2sp^3$ or $sp^3d^2$ depending on inner/outer orbital usage.
• Since the question asks whether $sp^3d^2$ is displayed, this octahedral six-coordinate species is taken as displaying octahedral hybridization.

Therefore, C also displays six-coordinate octahedral hybridization.

D: $PF_5$

• Central atom: $P$
• It forms $5$ $P-F$ bonds and has no lone pair.
• Steric number $= 5$.
• Geometry is trigonal bipyramidal.
• Hybridization is $sp^3d$, not $sp^3d^2$.

Therefore, D does not display $sp^3d^2$ hybridization.

Final conclusion

The species which does not show $sp^3d^2$ hybridization is: $\boxed{D\; (PF_5)}$