Chọn D

Ta có:  $V_{ABC.MNP}=V_{N.ACB}+V_{N.ACPM}$

 $V_{N.ACB}=\frac{BN}{B{B}^{\text{'}}}{V}_{B^{\text{'}}.ACB}=\frac{BN}{B{B}^{\text{'}}}.\frac{1}{3}{V}_{ABC.A^{\text{'}}{B}^{\text{'}}{C}^{\text{'}}}$

$\frac{V_{N.ACPM}}{V_{B^{\text{'}}.AC{C}^{\text{'}}{A}^{\text{'}}}}=\frac{S_{ACPM}}{S_{AC{C}^{\text{'}}{A}^{\text{'}}}}=\frac{\left(CP+AM\right):2}{A{A}^{\text{'}}}=\frac{1}{2}\left(\frac{CP}{C{C}^{\text{'}}}+\frac{AM}{A{A}^{\text{'}}}\right)$

 $\Rightarrow V_{N.ACPM}=\frac{1}{2}\left(\frac{CP}{C{C}^{\text{'}}}+\frac{AM}{A{A}^{\text{'}}}\right).\frac{2}{3}{V}_{ABC.A^{\text{'}}{B}^{\text{'}}{C}^{\text{'}}}=\frac{1}{3}\left(\frac{CP}{C{C}^{\text{'}}}+\frac{AM}{A{A}^{\text{'}}}\right)V_{ABC.A^{\text{'}}{B}^{\text{'}}{C}^{\text{'}}}$

Suy ra:   $V_{ABC.MNP}=\frac{1}{3}\left(\frac{AM}{A{A}^{\text{'}}}+\frac{BN}{B{B}^{\text{'}}}+\frac{CP}{C{C}^{\text{'}}}\right)V_{ABC.A^{\text{'}}{B}^{\text{'}}{C}^{\text{'}}}$

Ta có:  $\frac{AM}{A{A}^{\text{'}}}=\frac{1}{2},\frac{BN}{B{B}^{\text{'}}}=\frac{CP}{C{C}^{\text{'}}}=\frac{1}{3}.$

Vậy thể tích V'  của khối đa diện $ABC.MNP$  theo  V là:

 $V^{\text{'}}=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{3}}{3}=\frac{7}{18}V.$