Chọn phương án A.

$\blacksquare$ Trong mặt phẳng $(BCD)$, gọi $I=MN\cap CD$:

Kẻ đoạn thẳng $MM'\parallel BD$, khi đó $\dfrac{MM'}{BD}=\dfrac{CM}{CB}=\dfrac{2}{3}$.

Ngoài ra, vì $\dfrac{ND}{BD}=\dfrac{1}{3}$ nên $MM'=2ND$.

Tam giác $IMM'$ có $MM'\parallel ND$ và $MM'=2ND$ nên $ND$ là đường trung bình.

Do đó $DI=DM'=\dfrac{1}{3}DC$.

$\blacksquare$ Trong mặt phẳng $(ACD)$, gọi $Q=IP\cap AD$:

Kẻ đoạn thẳng $DD'\parallel IP$, khi đó $\dfrac{CD'}{CP}=\dfrac{CD}{CI}=\dfrac{3}{4}$.
Suy ra $\dfrac{AP}{AD'}=\dfrac{\dfrac{1}{2}AC}{\dfrac{5}{8}AC}=\dfrac{4}{5}$.

Tam giác $AD'D$ có $PQ\parallel D'D$ nên $\dfrac{AQ}{AD}=\dfrac{AP}{AD'}=\dfrac{4}{5}$.

$\blacksquare$ Trong tam giác $BCD$, gọi $h_C$ là chiều cao kẻ từ đỉnh $C$, ta có $$\begin{aligned}
S_{CNB}&=\dfrac{1}{2}\cdot BN\cdot h_C=\dfrac{1}{2}\cdot\dfrac{2}{3}BD\cdot h_C=\dfrac{2}{3}S_{CBD}\\
\Rightarrow V_{A.CNB}&=\dfrac{1}{3}\cdot S_{CNB}\cdot\mathrm{d}\big(A,(CNB)\big)\\
&=\dfrac{1}{3}\cdot\dfrac{2}{3}\cdot S_{CBD}\cdot\mathrm{d}\big(A,(CNB)\big)=\dfrac{2}{3}V_{A.CBD}\\
\Rightarrow V_{A.CND}&=V_{A.CBD}-V_{A.CNB}=\dfrac{1}{3}V_{A.CBD}.
\end{aligned}$$

$\blacksquare$ Xét khối chóp $C.NBA$ ta có $$\dfrac{V_{C.MNP}}{V_{C.BNA}}=\dfrac{CM}{CB}\cdot\dfrac{CN}{CN}\cdot\dfrac{CP}{CA}=\dfrac{2}{3}\cdot1\cdot\dfrac{1}{2}=\dfrac{1}{3}$$
Suy ra $V_{C.MNP}=\dfrac{1}{3}V_{C.BNA}=\dfrac{1}{3}\cdot\dfrac{2}{3}V_{A.CBD}=\dfrac{2}{9}V_{A.CBD}$.

Do đó $V_{N.ABMP}=\dfrac{2}{3}V_{C.BNA}=\dfrac{2}{3}\cdot\dfrac{2}{3}V_{A.CBD}=\dfrac{4}{9}V_{A.CBD}$. (1)

$\blacksquare$ Xét khối chóp $A.NCD$ ta có $$\dfrac{V_{A.NQP}}{V_{A.NCD}}=\dfrac{AN}{AN}\cdot\dfrac{AP}{AC}\cdot\dfrac{AQ}{AD}=1\cdot\dfrac{1}{2}\cdot\dfrac{4}{5}=\dfrac{2}{5}$$
Suy ra $V_{A.NQP}=\dfrac{2}{5}V_{A.NCD}=\dfrac{2}{5}\cdot\dfrac{1}{3}V_{A.CBD}=\dfrac{2}{15}V_{A.CBD}$. (2)

Từ (1) và (2) suy ra $V_1=V_{N.ABMP}+V_{A.NQP}=\dfrac{26}{45}V_{A.CBD}$.

Suy ra $V_2=V_{A.CBD}-V_1=\dfrac{19}{45}V_{A.CBD}$.

Vậy $\dfrac{V_1}{V_2}=\dfrac{26}{19}$.