Here, $\left|z-2\right|\le 1$ Put $z=x+iy$

${\left(x-2\right)}^2+y^2\le 1$

Also, $z\left(1+i\right)+\overline{z}\left(1-i\right)\ge 4$

Gives $x-y\ge 2$

Let point on circle be $A\left(2+\cos \theta ,\sin \theta \right)$

$\theta \in \left[-\frac{3\pi }{4},\frac{\pi }{4}\right]$

Let $P$ be $\left(\frac{5}{2},0\right)$

So, 

${\left|z-\frac{5}{2}\right|}^2={\left(\mathrm{AP}\right)}^2={\left(2+\cos \theta -\frac{5}{2}\right)}^2+{\sin }^2\theta$

$={\cos }^2\theta -\cos \theta +\frac{1}{4}+{\sin }^2\theta$

$=\frac{5}{4}-\cos \theta$

For ${\left(\mathrm{AP}\right)}^2$ to be maximum $\theta =-\frac{3\pi }{4}$

${\left(\mathrm{AP}\right)}^2=\frac{5}{4}+\frac{1}{\sqrt{2}}=\frac{5\sqrt{2}+4}{4\sqrt{2}}=\frac{5+2\sqrt{2}}{4}$