Statement-1: Alkali metals dissolve in liquid ammonia to give blue solutions. 

Statement-2: Alkali metals in liquid ammonia give solvated species of the type ${\left[\mathrm{M}{\left({\mathrm{NH}}_3\right)}_{\mathrm{n}}\right]}^{+}(\mathrm{M}=$alkali metals).

In liquid ammonia, the alkali metals dissolve without any hydrogen evolution. The diluted solutions are blue in colour.

${\mathrm{Na}}^{+}+(\mathrm{x}+\mathrm{y}){\mathrm{NH}}_3\to {\left[\mathrm{Na}(\mathrm{NH}3{)}_{\mathrm{x}}\right]}^{+}+\underset{\text{ Ammoniated electrons }}{\left[\mathrm{e}{\left({\mathrm{NH}}_3\right]}_{\mathrm{y}}\right.}$

Its blue tint turns bronze when heated. The colour and other characteristics are due to an ammoniated electron, not a solvated species of the kind ${\left[\mathrm{M}{\left({\mathrm{NH}}_3\right)}_{\mathrm{n}}\right]}^{+}(\mathrm{M}=$alkali metals).. The solutions have potent reducing characteristics and are effective electrical conductors. The remedies have a paramagnetic composition.
Therefore, both Statements I and II are accurate, but Statement II is not the proper justification for Statement I.
Therefore, choice B is right.