Step-by-step Solution:

Given line: \[ \frac{x}{4} + \frac{y}{2} = 1 \] Point (A) (x-intercept) Set (y = 0): \[ \frac{x}{4} = 1 \Rightarrow x = 4 \] \[ A = (4,0) \] Point (B) (y-intercept) Set (x = 0): \[ \frac{y}{2} = 1 \Rightarrow y = 2 \] \[ B = (0,2) \] Midpoint (M) of (AB) \[ M = \left(\frac{4+0}{2}, \frac{0+2}{2}\right) \] \[ M = (2,1) \] Step 2: Image (M') of (M) across the line (x+y=1) Rewrite line as: \[ x + y - 1 = 0 \] Using reflection formula for point ((x_1,y_1)) across line (ax+by+c=0): \[ \frac{h-x_1}{a} = \frac{k-y_1}{b} = -2 \frac{ax_1+by_1+c}{a^2+b^2} \] Here: \[ a=1,; b=1,; c=-1 \] \[ (x_1,y_1) = (2,1) \] Compute: \[ ax_1 + by_1 + c = 2 + 1 - 1 = 2 \] \[ a^2 + b^2 = 1 + 1 = 2 \] \[ -2 \cdot \frac{2}{2} = -2 \] Thus: \[ h - 2 = -2 \Rightarrow h = 0 \] \[ k - 1 = -2 \Rightarrow k = -1 \] \[ M' = (0,-1) \] Step 3: Coordinates of (P) Since (\triangle ABP) is isosceles with (AP = BP), point (P) lies on the perpendicular bisector of (AB). Also, (P) lies on line: \[ x+y=1 \] Perpendicular Bisector of (AB) Slope of (AB): \[ m_{AB} = \frac{2-0}{0-4} = -\frac{1}{2} \] Perpendicular slope: \[ m_\perp = 2 \] Equation passing through midpoint (M(2,1)): \[ y - 1 = 2(x - 2) \] \[ y = 2x - 3 \] \[ 2x - y = 3 \] Solve the System \[ x + y = 1 \] \[ 2x - y = 3 \] Add equations: \[ 3x = 4 \] \[ x = \frac{4}{3} \] Substitute: \[ \frac{4}{3} + y = 1 \] \[ y = -\frac{1}{3} \] \[ P = \left(\frac{4}{3}, -\frac{1}{3}\right) \] Step 4: Distance between (M') and (P) \[ d = \sqrt{\left(\frac{4}{3}-0\right)^2 + \left(-\frac{1}{3} - (-1)\right)^2} \] \[ d = \sqrt{\left(\frac{4}{3}\right)^2 + \left(\frac{2}{3}\right)^2} \] \[ d = \sqrt{\frac{16}{9} + \frac{4}{9}} \] \[ d = \sqrt{\frac{20}{9}} \] \[ d = \frac{\sqrt{20}}{3} \] \[ {d = \frac{2\sqrt{5}}{3}} \]