The incorrect geometry is represented by:

The geometry of a molecule is decided by the steric number (total of σ-bonds + lone pairs) around the central atom according to VSEPR theory.
Steric number = 3 → $$sp^2$$ hybridisation → trigonal planar
Steric number = 4 → $$sp^3$$ hybridisation → tetrahedral arrangement of electron pairs; with one lone pair it becomes trigonal pyramidal, with two lone pairs it becomes bent.
Steric number = 5 → $$sp^3d$$ hybridisation → trigonal bipyramidal.

Case A : NF3
Central atom N has 5 valence electrons. It forms three N-F σ-bonds and retains one lone pair.
Steric number = 3 (bonds) + 1 (lone pair) = 4 → $$sp^3$$ hybridisation, tetrahedral e-pair geometry.
Because one position is occupied by a lone pair, the observed molecular geometry is trigonal pyramidal, not trigonal planar. Hence the geometry mentioned in Option A is incorrect.

Case B : BF3
Boron forms three σ-bonds and has no lone pair.
Steric number = 3 → $$sp^2$$ hybridisation → trigonal planar geometry. Option B states “trigonal planar”, which is correct.

Case C : AsF5
Arsenic forms five σ-bonds with no lone pair.
Steric number = 5 → $$sp^3d$$ hybridisation → trigonal bipyramidal geometry. Option C is correct.

Case D : H2O
Oxygen forms two σ-bonds and has two lone pairs.
Steric number = 2 + 2 = 4 → $$sp^3$$ hybridisation; with two lone pairs the molecular shape is bent (angular). Option D is correct.

Therefore, the only mismatched pair of molecule and geometry is

Option A which is: NF3 - trigonal planar.