Đáp án A

Chọn A

Trong \((SBD)\), đường thẳng qua \(M\) và song song với \(BD\) cắt \(SD\) tại \(P\).
Trong \((ABCD)\), \(AC\) cắt \(BD\) tại \(O\).
Trong \((SBD)\), \(SO\) cắt \(MP\) tại \(G\).
Trong \((SAC)\), \(AG\) cắt \(SC\) tại \(N\).
Trong \(\left( {SAB} \right)\), \(SE\) cắt \(AM\) tại \(F\).
Ta có \({V_{E.AMNP}} = \dfrac{1}{2}{V_{B.AMNP}}\) và \({V_{B.AMNP}} = \dfrac{1}{4}{V_{S.AMNP}}\) \( \Rightarrow \,\,\,{V_{E.AMNP}} = \dfrac{1}{8}{V_{S.AMNP}}\)
Theo định lý Thales trong tam giác \(SBD\): \(\dfrac{{SP}}{{SD}} = \dfrac{{SM}}{{SB}} = \dfrac{4}{5}\).
Mặt khác: \(\dfrac{{SA}}{{SA}} + \dfrac{{SC}}{{SN}} = \dfrac{{SB}}{{SM}} + \dfrac{{SD}}{{SP}} \Rightarrow \,\,\,1 + \dfrac{{SC}}{{SN}} = \dfrac{5}{4} + \dfrac{5}{4} \Rightarrow \,\,\,\dfrac{{SC}}{{SN}} = \dfrac{3}{2} \Rightarrow \,\,\,\dfrac{{SN}}{{SC}} = \dfrac{2}{3}\).
Ta có: \(\dfrac{{{V_{S.AMN}}}}{{{V_{S.ABCD}}}} = \dfrac{1}{2}.\dfrac{{{V_{S.AMN}}}}{{{V_{S.ABC}}}} = 2.\dfrac{{SA}}{{SA}}.\dfrac{{SM}}{{SB}}.\dfrac{{SN}}{{SC}} = \dfrac{1}{2}.1.\dfrac{4}{5}.\dfrac{2}{3} = \dfrac{4}{{15}}\,\, \Rightarrow \,\,\,\,{V_{S.AMN}} = \dfrac{4}{{15}}{V_{S.ABCD}}\)
\(\dfrac{{{V_{S.APN}}}}{{{V_{S.ABCD}}}} = \dfrac{1}{2}.\dfrac{{{V_{S.APN}}}}{{{V_{S.ADC}}}} = 2.\dfrac{{SA}}{{SA}}.\dfrac{{SP}}{{SD}}.\dfrac{{SN}}{{SC}} = \dfrac{1}{2}.1.\dfrac{4}{5}.\dfrac{2}{3} = \dfrac{4}{{15}}\,\, \Rightarrow \,\,\,\,{V_{S.APN}} = \dfrac{4}{{15}}{V_{S.ABCD}}\)
Suy ra \({V_{S.AMNP}} = {V_{S.AMN}} + {V_{S.APN}} = \dfrac{8}{{15}}{V_{S.ABCD}}\).
Mà \({V_{S.ABCD}} = \dfrac{1}{3}.2a.{a^2} = \dfrac{2}{3}{a^3}\)
\( \Rightarrow \,\,\,{V_{S.AMNP}} = \dfrac{8}{{15}}.\dfrac{2}{3}{a^3} = \dfrac{{16{a^3}}}{{45}} \Rightarrow \,\,\,{V_{E.AMNP}} = \dfrac{1}{8}.\dfrac{{16{a^3}}}{{45}} = \dfrac{{2{a^3}}}{{45}}\).