The geometry around boron in the product ' B ' formed from the...

$2{BF}_{3} + 6NaH \to \mathop{{B}_{2}{H}_{6}}_{\left(A\right)} + 6NaF$

In this reaction hydride ions replace the fluoride ions. and diborane is formed, which is a Lewis acid due to incomplete octet. It breaks symmetrically with bases like tertiary amines.

${B}_{2}{H}_{6} + 2{Me}_{3}N \to \mathop{2{\left({\text{ Me }}_{3}N\right)}^{+}}$_{\left(B\right)}{\left({BH}_{3}\right)}^{-}

In this product geometry around Boron will be tetrahedral. And it will have ${sp}^{3}$ hybridisation.