Gọi \(x\) là độ dài cạnh đáy của chóp đều \(S.ABCD\).

Gọi \(O = AC \cap BD \Rightarrow SO \bot \left( {ABCD} \right)\).

Ta có:

\(\left\{ \begin{array}{l}BD \bot AC\,\,\left( {gt} \right)\\BD \bot SO\,\,\left( {SO \bot \left( {ABCD} \right)} \right)\end{array} \right. \Rightarrow BD \bot \left( {SAC} \right) \Rightarrow BD \bot SC\).

Trong \(\left( {SBC} \right)\) kẻ \(BH \bot SC\,\,\left( {H \in SC} \right)\) ta có

\(\left\{ \begin{array}{l}BH \bot SC\\BD \bot SC\,\,\left( {cmt} \right)\end{array} \right. \Rightarrow SC \bot \left( {BDH} \right) \Rightarrow SC \bot DH\).

Ta có : \(\left\{ \begin{array}{l}\left( {SBC} \right) \cap \left( {SCD} \right) = SC\\\left( {SBC} \right) \supset BH \bot SC\\\left( {SCD} \right) \supset DH \bot SC\end{array} \right. \Rightarrow \angle \left( {\left( {SBC} \right);\left( {SCD} \right)} \right) = \angle \left( {BH;DH} \right) \Rightarrow \left[ \begin{array}{l}\cos \angle BHD = \frac{1}{{10}}\\\cos \angle BHD =  - \frac{1}{{10}}\end{array} \right.\).

Dễ dàng chứng minh được \(\Delta BHC = \Delta DHC \Rightarrow HB = HD \Rightarrow \Delta HBD\) cân tại \(H\).

Xét tam giác \(SBC\) ta có : \(\cos \angle C = \frac{{B{C^2} + S{C^2} - S{B^2}}}{{2BC.SC}} = \frac{{{x^2}}}{{2x.\sqrt {11} a}} = \frac{{x\sqrt {11} }}{{22a}}\)

\( \Rightarrow HC = BC.\cos  \angle C = \frac{{{x^2}\sqrt {11} }}{{22a}}\).

\( \Rightarrow HB = \sqrt {B{C^2} - H{C^2}}  = \sqrt {{x^2} - \frac{{{x^4}}}{{44{a^2}}}}  = \frac{{x\sqrt {{a^2} - {x^2}} }}{{2a\sqrt {11} }} = HD\)

 Xét tam giác \(BDH\) có :

\(\cos \angle BHD = \frac{{H{B^2} + H{D^2} - B{D^2}}}{{2HB.HD}} = \frac{{2{x^2} - \frac{{{x^4}}}{{22{a^2}}} - 2{x^2}}}{{2\left( {{x^2} - \frac{{{x^4}}}{{44{a^2}}}} \right)}} = \frac{{2{x^2} - \frac{{{x^4}}}{{22{a^2}}} - 2{x^2}}}{{2{x^2} - \frac{{{x^4}}}{{22{a^2}}}}} = 1 - \frac{{44{x^2}{a^2}}}{{44{x^2}{a^2} - {x^4}}}\)

TH1: \(\cos \angle BHD = \frac{1}{{10}} \Leftrightarrow 1 - \frac{{44{x^2}{a^2}}}{{44{x^2}{a^2} - {x^4}}} = \frac{1}{{10}} \Leftrightarrow \frac{{44{x^2}{a^2}}}{{44{x^2}{a^2} - {x^4}}} = \frac{9}{{10}}\) 

\( \Leftrightarrow 440{x^2}{a^2} = 396{x^2}{a^2} - 9{x^4} \Leftrightarrow 9{x^4} =  - 44{x^2}{a^2}\) (vô nghiệm)

TH2: \(\cos \angle BHD =  - \frac{1}{{10}} \Leftrightarrow 1 - \frac{{44{x^2}{a^2}}}{{44{x^2}{a^2} - {x^4}}} =  - \frac{1}{{10}} \Leftrightarrow \frac{{44{x^2}{a^2}}}{{44{x^2}{a^2} - {x^4}}} = \frac{{11}}{{10}}\)

\(\begin{array}{l} \Leftrightarrow 440{x^2}{a^2} = 484{x^2}{a^2} - 11{x^4} \Leftrightarrow 11{x^4} = 44{x^2}{a^2} \Leftrightarrow {x^2} = 4{a^2} \Leftrightarrow x = 2a\\ \Rightarrow OA = \frac{1}{2}AC = \frac{1}{2}.2a\sqrt 2  = a\sqrt 2 \end{array}\).

Xét tam giác vuông \(SOA\) có : \(SO = \sqrt {S{A^2} - O{A^2}}  = \sqrt {11{a^2} - 2{a^2}}  = 3a\).

Vậy \({V_{S.ABCD}} = \frac{1}{3}SO.{S_{ABCD}} = \frac{1}{3}.3a.{\left( {2a} \right)^2} = 4{a^3}\).

Chọn C.