The geometry around boron in the product 'B' formed from the following reaction is
$$BF_3 + NaH \xrightarrow{450K} A + NaF$$
$$A + NMe_3 \rightarrow B$$

Step 1: Formation of Product 'A'

When boron trifluoride ($$\text{BF}_3$$) reacts with sodium hydride ($$\text{NaH}$$) at around $$450\text{ K}$$, it undergoes a reduction reaction to form diborane ($$\text{B}_2\text{H}_6$$) along with sodium fluoride ($$\text{NaF}$$):

$$2\text{BF}_3 + 6\text{NaH} \xrightarrow{450\text{K}} \text{B}_2\text{H}_6 + 6\text{NaF}$$

Therefore, product 'A' is diborane, $$\text{B}_2\text{H}_6$$.

Step 2: Formation of Product 'B'

Diborane ($$\text{B}_2\text{H}_6$$) is an electron-deficient molecule that easily undergoes cleavage by Lewis bases. When treated with a relatively small, bulky tertiary amine like trimethylamine ($$\text{NMe}_3$$), it undergoes symmetric cleavage to yield a stable Lewis acid-base adduct:

$$\text{B}_2\text{H}_6 + 2\text{NMe}_3 \rightarrow 2\text{BH}_3\cdot\text{NMe}_3$$

Therefore, product 'B' is the monomeric adduct, $$\text{BH}_3\cdot\text{NMe}_3$$.

Hybridization and Geometry Around Boron in Product 'B':

In the adduct $$\text{BH}_3\cdot\text{NMe}_3$$, the boron atom accepts a lone pair of electrons from the nitrogen atom of the trimethylamine molecule via a coordinate (dative) bond.

• The central Boron atom forms 3 covalent bonds with hydrogen atoms ($$\text{B--H}$$) and 1 coordinate bond with the nitrogen atom ($$\text{B}\leftarrow\text{N}$$).
• This gives a total of 4 bond pairs and 0 lone pairs on the boron atom.
• According to the VSEPR theory, a steric number of 4 corresponds to $$\text{sp}^3$$ hybridization.

The spatial arrangement corresponding to $$\text{sp}^3$$ hybridization with zero lone pairs is completely symmetrical, giving it a tetrahedral geometry.

Conclusion:

The coordination of the lone pair from trimethylamine changes the geometry around the boron atom from trigonal planar (in $$\text{BH}_3$$ fragments) to tetrahedral in the final complex.

Answer: Option B — tetrahedral