1. Let $z=x+iy$ with $x,y\in\mathbb{R}$.

   Then $\bar z=x-iy$.

2. Convert each condition into Cartesian form.

   (i) Circle condition

   $|z-1|\le 2$ means $|(x-1)+iy|\le 2 \implies (x-1)^2+y^2\le 4.$

   So this is the disk centered at $(1,0)$ with radius $2$.

   (ii) Linear condition

   We compute: $z+\bar z=(x+iy)+(x-iy)=2x,$ and $i(z-\bar z)=i[(x+iy)-(x-iy)]=i(2iy)=-2y.$

   Therefore $(z+\bar z)+i(z-\bar z)=2x-2y.$

   Given $2x-2y\le 2 \implies x-y\le 1 \implies y\ge x-1.$

   (iii) Imaginary part condition

   $\operatorname{Im}(z)\ge 0 \implies y\ge 0.$

3. Describe the required region.

   We need the part of the disk $(x-1)^2+y^2\le 4$ satisfying both $y\ge 0 \quad \text{and} \quad y\ge x-1.$

   So inside the circle, we want the region lying above the $x$-axis and above the line $y=x-1$.

4. Understand the geometry of the line relative to the circle.

   The circle has center $(1,0)$ and radius $2$.

   The line is $y=x-1 \iff x-y-1=0.$

   Distance from center $(1,0)$ to this line is $\frac{|1-0-1|}{\sqrt{1^2+(-1)^2}}=0.$

   Hence the line passes through the center of the circle. Therefore, inside the circle it cuts the disk into two equal halves.

   So:

   • area of the full disk $=\pi(2)^2=4\pi$,
   • area of the part of the disk with $y\ge x-1$ $=2\pi$.

5. Now impose also $y\ge 0$.

   The upper semicircle of the given disk has area $\frac{1}{2}\cdot 4\pi=2\pi.$

   We need the intersection of:

   • upper semicircle $y\ge 0$,
   • half-disk $y\ge x-1$.

   Equivalently, from the upper semicircle, remove the portion where $y\ge 0 \quad \text{but} \quad y<x-1.$

   Inside the disk, the line $y=x-1$ meets the circle at endpoints of a diameter. Since the center is on the $x$-axis, in the upper semicircle this line cuts off a sector of angle $\frac{\pi}{2}$ between radii toward the intersection points:

   • rightmost point of circle: $(3,0)$,
   • upper-left intersection with the line: solving with center-angle gives point $(1-\sqrt2, -\sqrt2)?$ Better to use direction of the line.

   A cleaner way: translate center to origin by setting $X=x-1,\quad Y=y.$ Then circle becomes $X^2+Y^2\le 4,$ and inequalities become $Y\ge 0,\qquad Y\ge X.$

   In polar coordinates about the center,

   $$$$

\le 2.$$

Then $Y\ge 0 \implies \sin\theta\ge 0 \implies 0\le \theta\le \pi,$ and $Y\ge X \implies \sin\theta\ge \cos\theta \implies \sin\theta-\cos\theta\ge 0.$

This gives $\sqrt2\sin\left(\theta-\frac{\pi}{4}\right)\ge 0,$ so within $[0,\pi]$ we get $\frac{\pi}{4}\le \theta\le \pi.$

Hence the required region is a sector of radius $2$ and angle $\pi-\frac{\pi}{4}=\frac{3\pi}{4}.$

6. Compute the area.

   Area of sector: $\frac{1}{2}r^2\theta=\frac{1}{2}\cdot 4\cdot \frac{3\pi}{4}=\frac{3\pi}{2}.$

7. Check options.

   $\frac{3\pi}{2}$ matches Option B.

Final Answer: $\boxed{\frac{3\pi}{2}}$