To determine if a pair of species is isostructural, we need to compare their shapes and hybridization. This involves calculating the steric number (number of bond pairs + lone pairs) for the central atom, which then determines the hybridization and molecular geometry.

Step 1: Analyze Option A - ${\mathrm{NF}}_3$ and ${\mathrm{BF}}_3$

• For ${\mathrm{NF}}_3$: Nitrogen (central atom) has 5 valence electrons. It forms 3 single bonds with F atoms and has 1 lone pair. Steric number = 3 (bond pairs) + 1 (lone pair) = 4. Hybridization = ${\mathrm{sp}}^3$. Shape = Pyramidal.
• For ${\mathrm{BF}}_3$: Boron (central atom) has 3 valence electrons. It forms 3 single bonds with F atoms and has 0 lone pairs. Steric number = 3 (bond pairs) + 0 (lone pairs) = 3. Hybridization = ${\mathrm{sp}}^2$. Shape = Trigonal planar.
• Since their shapes and hybridizations are different, they are not isostructural.

Step 2: Analyze Option B - ${\mathrm{BF}}_4^{-}$ and ${\mathrm{NH}}_4^{+}$

• For ${\mathrm{BF}}_4^{-}$: Boron (central atom) has 3 valence electrons. It forms 4 single bonds with F atoms and has a -1 charge, meaning it accepts one electron. So, 3 (valence e-) + 1 (from charge) = 4 electrons for bonding. It forms 4 single bonds with F atoms and has 0 lone pairs. Steric number = 4 (bond pairs) + 0 (lone pairs) = 4. Hybridization = ${\mathrm{sp}}^3$. Shape = Tetrahedral.
• For ${\mathrm{NH}}_4^{+}$: Nitrogen (central atom) has 5 valence electrons. It forms 4 single bonds with H atoms and has a +1 charge, meaning it loses one electron. So, 5 (valence e-) - 1 (from charge) = 4 electrons for bonding. It forms 4 single bonds with H atoms and has 0 lone pairs. Steric number = 4 (bond pairs) + 0 (lone pairs) = 4. Hybridization = ${\mathrm{sp}}^3$. Shape = Tetrahedral.
• Since their shapes and hybridizations are the same, they are isostructural.

Step 3: Analyze Option C - ${\mathrm{BCl}}_3$ and ${\mathrm{BrCl}}_3$

• For ${\mathrm{BCl}}_3$: Boron (central atom) has 3 valence electrons. It forms 3 single bonds with Cl atoms and has 0 lone pairs. Steric number = 3. Hybridization = ${\mathrm{sp}}^2$. Shape = Trigonal planar.
• For ${\mathrm{BrCl}}_3$: Bromine (central atom) has 7 valence electrons. It forms 3 single bonds with Cl atoms and has 2 lone pairs (7 - 3 = 4 electrons, so 2 lone pairs). Steric number = 3 (bond pairs) + 2 (lone pairs) = 5. Hybridization = ${\mathrm{sp}}^3\mathrm{d}$. Shape = T-shaped.
• Since their shapes and hybridizations are different, they are not isostructural.

Step 4: Analyze Option D - ${\mathrm{NH}}_3$ and ${\mathrm{NO}}_3^{-}$

• For ${\mathrm{NH}}_3$: Nitrogen (central atom) has 5 valence electrons. It forms 3 single bonds with H atoms and has 1 lone pair. Steric number = 4. Hybridization = ${\mathrm{sp}}^3$. Shape = Pyramidal.
• For ${\mathrm{NO}}_3^{-}$: Nitrogen (central atom) has 5 valence electrons. It forms 3 bonds (one double, two single) with O atoms and has a -1 charge. The total number of valence electrons for ${\mathrm{NO}}_3^{-}$ is 5 (N) + 3*6 (O) + 1 (charge) = 24. The central N forms 3 bonds with O atoms and has 0 lone pairs. Steric number = 3. Hybridization = ${\mathrm{sp}}^2$. Shape = Trigonal planar.
• Since their shapes and hybridizations are different, they are not isostructural.

Correct Answer: (B)