Let $$\vec{a}$$ and $$\vec{b}$$ be the vectors along the diagonal of a parallelogram having area $$2\sqrt{2}$$. Let the angle between $$\vec{a}$$ and $$\vec{b}$$ be acute. $$|\vec{a}| = 1$$ and $$|\vec{a} \cdot \vec{b}| = |\vec{a} \times \vec{b}|$$. If $$\vec{c} = 2\sqrt{2}(\vec{a} \times \vec{b}) - 2\vec{b}$$, then an angle between $$\vec{b}$$ and $$\vec{c}$$ is

Let the diagonals of the parallelogram be

$$\vec a \quad \text{and} \quad \vec b$$

Area of a parallelogram in terms of diagonals is

$$\text{Area}=\frac12|\vec a\times \vec b|$$

Given,

$$\frac12|\vec a\times \vec b|=2\sqrt2$$

Hence,

$$|\vec a\times \vec b|=4\sqrt2$$

Now,

$$|\vec a|=1$$

and

$$|\vec a\cdot \vec b|=|\vec a\times \vec b|$$

Let the angle between $$\vec a$$ and $$\vec b$$ be $$\theta$$.

Then,

$$|\vec a||\vec b|\cos\theta=|\vec a||\vec b|\sin\theta$$

$$\tan\theta=1$$

Since the angle is acute,

$$\theta=\frac{\pi}{4}$$

Also,

$$|\vec a\times \vec b| = |\vec a||\vec b|\sin\frac{\pi}{4} $$

$$4\sqrt2=1\cdot |\vec b|\cdot \frac1{\sqrt2}$$

$$|\vec b|=8$$

Given,

$$\vec c=2\sqrt2(\vec a\times \vec b)-2\vec b$$

Now,

$$\vec b\cdot(\vec a\times \vec b)=0$$

Therefore,

$$\vec b\cdot\vec c = -2|\vec b|^2 = -2(64) = -128 $$

Also,

$$|2\sqrt2(\vec a\times \vec b)|=16$$

and

$$|-2\vec b|=16$$

Since these two vectors are perpendicular,

$$|\vec c| = \sqrt{16^2+16^2} = 16\sqrt2$$

Let the angle between $$\vec b$$ and $$\vec c$$ be $$\alpha$$.

Then,

$$\cos\alpha = \frac{\vec b\cdot\vec c}{|\vec b||\vec c|} = \frac{-128}{8\cdot16\sqrt2} = -\frac1{\sqrt2} $$

Hence,

$$\alpha=\frac{3\pi}{4}$$

Therefore, the correct answer is

$$\boxed{\text{Option D }\frac{3\pi}{4}}$$