Trong \(\left( {BCD} \right)\) gọi \(E = MN \cap CD\).

Trong \(\left( {ACD} \right)\) gọi \(Q = AD \cap PE\).

Khi đó thiết diện của hình chóp khi cắt bởi mặt phẳng \(\left( {MNP} \right)\) là tứ giác \(MNQP\).

Áp dụng định lí Menelaus trong tam giác BCD ta có:

\(\dfrac{{MB}}{{MC}}.\dfrac{{EC}}{{ED}}.\dfrac{{ND}}{{NB}} = 1 \Rightarrow \dfrac{1}{2}.\dfrac{{EC}}{{ED}}.\dfrac{1}{2} = 1 \Leftrightarrow \dfrac{{EC}}{{ED}} = 4\)

Áp dụng định lí Menelaus trong tam giác ACD ta có:

\(\dfrac{{PA}}{{PC}}.\dfrac{{EC}}{{ED}}.\dfrac{{QD}}{{QA}} = 1 \Rightarrow 1.4.\dfrac{{QD}}{{QA}} = 1 \Rightarrow \dfrac{{QD}}{{QA}} = \dfrac{1}{4}\) Ta có: \({V_{ABMNQ}} = {V_{ABMN}} + {V_{AMNP}} + {V_{ANPQ}}\)

\(\begin{array}{l} + )\,\,\dfrac{{{S_{BMN}}}}{{{S_{BCD}}}} = \dfrac{{BM}}{{BC}}.\dfrac{{BN}}{{BD}} = \dfrac{1}{3}.\dfrac{2}{3} = \dfrac{2}{9} \Rightarrow \dfrac{{{V_{ABMN}}}}{{{V_{ABCD}}}} = \dfrac{2}{9}\\ + )\,\,\dfrac{{{V_{AMNP}}}}{{{V_{AMNC}}}} = \dfrac{{AP}}{{AC}} = \dfrac{1}{2} \Rightarrow {V_{AMNP}} = \dfrac{1}{2}{V_{AMNC}}\\\,\,\,\,\,\,\dfrac{{{S_{NMC}}}}{{{S_{DBC}}}} = \dfrac{{d\left( {N;BC} \right).MC}}{{d\left( {D ;BC} \right).BC}} = \dfrac{{NB}}{{DB}}.\dfrac{{MC}}{{BC}} = \dfrac{2}{3}.\dfrac{2}{3} = \dfrac{4}{9}\\ \Rightarrow \dfrac{{{V_{AMNC}}}}{{{V_{ABCD}}}} = \dfrac{4}{9} \Rightarrow {V_{AMNP}} = \dfrac{2}{9}{V_{ABCD}}\\ + )\,\,\dfrac{{{V_{APQN}}}}{{{V_{ACDN}}}} = \dfrac{{AP}}{{AC}}.\dfrac{{AQ}}{{AD}} = \dfrac{1}{2}.\dfrac{4}{5} = \dfrac{2}{5} \Rightarrow {V_{APQN}} = \dfrac{2}{5}{V_{ACDN}}\\\,\,\,\,\,\,\dfrac{{{S_{CND}}}}{{{S_{CBD}}}} = \dfrac{{DN}}{{DB}} = \dfrac{1}{3} \Rightarrow \dfrac{{{V_{ACDN}}}}{{{V_{ABCD}}}} = \dfrac{1}{3} \Rightarrow {V_{APQN}} = \dfrac{2}{{15}}{V_{ABCD}}\end{array}\)

\( \Rightarrow {V_{ABMNQ}} = {V_{ABMN}} + {V_{AMNP}} + {V_{ANPQ}} = \dfrac{2}{9}{V_{ABCD}} + \dfrac{2}{9}{V_{ABCD}} + \dfrac{2}{{15}}{V_{ABCD}} = \dfrac{{26}}{{45}}{V_{ABCD}}\).

Gọi \({V_1} = {V_{ABMNQ}},\,\,{V_2}\) là thể tích phần còn lại \( \Rightarrow \dfrac{{{V_1}}}{{{V_2}}} = \dfrac{{26}}{{19}}\).

Chọn A.