$$|z_1 + z_2|^2 + |z_1 - z_2|^2$$ is equal to

Let $$z_1 = x_1 + i\,y_1$$ and $$z_2 = x_2 + i\,y_2$$, where $$x_1, x_2, y_1, y_2 \in \mathbb{R}$$.

Recall that for any complex number $$z = x + i\,y$$, its modulus is $$|z| = \sqrt{x^2 + y^2}$$, so $$|z|^2 = x^2 + y^2$$.

First compute $$|z_1 + z_2|^2$$.
$$z_1 + z_2 = (x_1 + x_2) + i\,(y_1 + y_2)$$
Therefore, $$|z_1 + z_2|^2 = (x_1 + x_2)^2 + (y_1 + y_2)^2 \quad -(1)$$

Next compute $$|z_1 - z_2|^2$$.
$$z_1 - z_2 = (x_1 - x_2) + i\,(y_1 - y_2)$$
Therefore, $$|z_1 - z_2|^2 = (x_1 - x_2)^2 + (y_1 - y_2)^2 \quad -(2)$$

Add the two results from $$-(1)$$ and $$-(2)$$:
$$|z_1 + z_2|^2 + |z_1 - z_2|^2$$ $$= \bigl[(x_1 + x_2)^2 + (y_1 + y_2)^2\bigr] + \bigl[(x_1 - x_2)^2 + (y_1 - y_2)^2\bigr]$$ $$= \bigl[(x_1^2 + 2x_1x_2 + x_2^2) + (y_1^2 + 2y_1y_2 + y_2^2)\bigr] + \bigl[(x_1^2 - 2x_1x_2 + x_2^2) + (y_1^2 - 2y_1y_2 + y_2^2)\bigr]$$

The cross terms $$\pm 2x_1x_2$$ and $$\pm 2y_1y_2$$ cancel out, leaving
$$= 2x_1^2 + 2x_2^2 + 2y_1^2 + 2y_2^2$$ $$= 2\bigl(x_1^2 + y_1^2 + x_2^2 + y_2^2\bigr)$$ $$= 2\bigl(|z_1|^2 + |z_2|^2\bigr)$$

Hence
$$|z_1 + z_2|^2 + |z_1 - z_2|^2 = 2\left(|z_1|^2 + |z_2|^2\right)$$

Therefore, the correct choice is Option B which is: $$2(|z_1|^2 + |z_2|^2)$$.