Let $$\mathbb{C}$$ be the set of all complex numbers. Let $$S_1 = \{z \in \mathbb{C} : |z - 2| \leq 1\}$$ and $$S_2 = \{z \in \mathbb{C} : z(1 + i) + \bar{z}(1 - i) \geq 4\}$$. Then, the maximum value of $$\left|z - \frac{5}{2}\right|^2$$ for $$z \in S_1 \cap S_2$$ is equal to:

Let us write the variable complex number as $$z = x + iy,\; x,y \in \mathbb{R}$$.

For the set $$S_1$$ we have the condition of a circle:

$$|z-2| \le 1 \; \Longrightarrow \; |(x+iy)-2| \le 1 \; \Longrightarrow\; (x-2)^2 + y^2 \le 1.$$

For the set $$S_2$$ start from the given expression and replace $$z$$ and $$\bar z$$:

$$z(1+i)+\bar z(1-i) \ge 4.$$

Because $$z = x+iy$$ and $$\bar z = x-iy,$$

$$z(1+i)= (x+iy)(1+i)=x(1+i)+iy(1+i) = (x - y) + i(x+y),$$

$$\bar z(1-i)= (x-iy)(1-i)=x(1-i)-iy(1-i) = (x - y) - i(x+y).$$

Adding both gives

$$\bigl[(x - y) + i(x+y)\bigr] + \bigl[(x - y) - i(x+y)\bigr] = 2(x-y).$$

Hence $$S_2$$ is the half-plane

$$2(x-y)\ge 4 \;\Longrightarrow\; x-y \ge 2 \;\Longrightarrow\; y \le x-2.$$

Therefore $$S_1\cap S_2$$ consists of the part of the closed disc $$(x-2)^2+y^2\le 1$$ that lies on or below the line $$y = x-2.$$

We have to maximise

$$\left|z-\frac52\right|^2 = (x-\tfrac52)^2 + y^2$$

subject to both constraints. The centre of the disc is $$C(2,0)$$ and the point from which we measure the distance is $$P\!\left(\frac52,0\right).$$ Since the distance is radially increasing, its maximum in the feasible region will occur on the boundary circle $$(x-2)^2+y^2 = 1.$$

Put $$x = 2 + \cos\theta,\; y = \sin\theta, \quad -\pi \le \theta \le \pi,$$ the standard parametrisation of this circle.

The half-plane condition $$y\le x-2$$ becomes

$$\sin\theta \le \cos\theta,$$

or

$$\cos\theta - \sin\theta \ge 0.$$

Using the identity $$\cos\theta - \sin\theta = \sqrt2\cos\!\left(\theta+\frac{\pi}{4}\right),$$ the inequality is $$\sqrt2\cos\!\left(\theta+\frac{\pi}{4}\right)\ge 0 \;\Longrightarrow\; -\frac{3\pi}{4}\le \theta \le \frac{\pi}{4}.$$

Now compute the required squared distance:

$$\left|z-\frac52\right|^2 = \bigl(2+\cos\theta-\tfrac52\bigr)^2 + (\sin\theta)^2 = (\cos\theta-\tfrac12)^2 + \sin^2\theta.$$

Expanding and using $$\sin^2\theta+\cos^2\theta=1,$$

$$\left|z-\frac52\right|^2 = \cos^2\theta - \cos\theta + \frac14 + \sin^2\theta = 1 - \cos\theta + \frac14 = \frac54 - \cos\theta.$$

Thus the problem reduces to maximising $$\frac54 - \cos\theta$$ over $$-\dfrac{3\pi}{4}\le\theta\le\dfrac{\pi}{4}.$$ Equivalently, we must minimise $$\cos\theta$$ on that interval. The cosine function attains its minimum at the left end-point $$\theta = -\dfrac{3\pi}{4},$$ where

$$\cos\!\left(-\frac{3\pi}{4}\right)= -\frac{\sqrt2}{2}.$$

Substituting this value,

$$\max\left|z-\frac52\right|^2 = \frac54 - \Bigl(-\frac{\sqrt2}{2}\Bigr) = \frac54 + \frac{\sqrt2}{2} = \frac{5 + 2\sqrt2}{4}.$$

Hence, the correct answer is Option D.