226 - Moment of force about different points

$M_A = 180 \, \text{ kg}\cdot \text{m}$       → answer
 
$M_B = -6(0.3)F_H = -6(0.3)(160)$

$M_B = -288 \, \text{ kg}\cdot \text{m}$       → answer
 
$M_C = -0.3F_V - 3(0.3)F_H = -0.3(120) - 3(0.3)(160)$

$M_C = -180 \, \text{ kg}\cdot \text{m}$       → answer
 
$M_D = 5(0.3)F_V - 6(0.3)F_H = 5(0.3)(120) - 6(0.3)(160)$

$M_C = -108 \, \text{ kg}\cdot \text{m}$       → answer

$M_A = 0$ (this means that point A is on the line of action of force P)       → answer
 
$M_B = 0.3P_V = 0.3(137.288)$

$M_B = 41.186 \, \text{ kg}\cdot \text{m}$       → answer
 
$M_C = 2(0.3)P_V + 3(0.3)P_H = 2(0.3)(137.288) + 3(0.3)(91.526)$

$M_C = 164.746 \, \text{ kg}\cdot \text{m}$       → answer

You can also resolve P to horizontal and vertical components at point E then take the moment of these components at point C. The answer would be the same. Try it.
 
$M_D = -4(0.3)P_V = -4(0.3)(137.288)$

$M_D = -164.746 \, \text{ kg}\cdot \text{m}$       → answer