The figure above is a Parallelogram, where, AB || CD and AD || BC. Also, BC = AD = 4 and AB = CD = 2.
Area of a Parallelogram = Base x Height

Drop a perpendicular from point C to line segment AD (let us name it is as point E)

Now, triangle CED is an Isosceles Right Triangle [45-45-90 triangle]
Ratio of sides in 45-45-90 triangle = $x : x : \sqrt{2}x$

opposite to 45 = $x$
opposite to 90 = $\sqrt{2}x$ = CD = 2

So, $x$ = $\sqrt{2}$
i.e. Height = $\sqrt{2}$

Therefore, Area of parallelogram = 4 x $\sqrt{2}$ = $4\sqrt{2}$

Hence, option B