Among the following, the pair in which the two species are not isostructural is :-

Find the number of bond pairs (bp) and lone pairs (lp) on the central atom for each species.
$IO_3^-$: I has 3 bp and 1 lp, so it is pyramidal ($sp^3$). $XeO_3$: Xe has 3 bp and 1 lp, so it is also pyramidal. They are isostructural.
$BH_4^-$: B has 4 bp, 0 lp, tetrahedral. $NH_4^+$: N has 4 bp, 0 lp, tetrahedral. They are isostructural.
$PF_6^-$: P has 6 bp, 0 lp, octahedral ($sp^3d^2$). $SF_6$: S has 6 bp, 0 lp, octahedral. They are isostructural.
$SiF_4$: Si has 4 bp and 0 lp, so it is tetrahedral ($sp^3$).
$SF_4$: S has 6 valence electrons; 4 are used in bonds with F, leaving 1 lone pair. So there are 4 bp + 1 lp ($sp^3d$), and the shape is see-saw.
So $SiF_4$ (tetrahedral) and $SF_4$ (see-saw) are not isostructural.