Phương pháp giải:

\(\left\{ \begin{array}{l}\left( P \right) \bot \left( \alpha  \right)\\\left( Q \right) \bot \left( \alpha  \right)\\\left( P \right) \cap \left( Q \right) = d\end{array} \right. \Rightarrow d \bot \left( \alpha  \right)\)

Lời giải chi tiết:

Ta có: \(\left\{ \begin{array}{l}\left( {SAC} \right) \bot \left( {ABCD} \right)\\\left( {SBD} \right) \bot \left( {ABCD} \right)\\\left( {SAC} \right) \cap \left( {SBD} \right) = SO\end{array} \right. \Rightarrow SO \bot \left( {ABCD} \right)\)

Dựng \(OH \bot AB,\,\,H \in AB;\,\,\,\,OK \bot SH,\,K \in SH\). Ta có: \(\left\{ \begin{array}{l}AB \bot OH\\AB \bot SO\end{array} \right. \Rightarrow AB \bot \left( {SOH} \right) \Rightarrow AB \bot OK\)

Mà \(OK \bot SH \Rightarrow OK \bot \left( {SAB} \right) \Rightarrow d\left( {O;\left( {SAB} \right)} \right) = OK = \dfrac{{a\sqrt 3 }}{4}\)

\(\Delta OAB\) vuông tại O, \(OH \bot AB \Rightarrow \dfrac{1}{{O{H^2}}} = \dfrac{1}{{O{A^2}}} + \dfrac{1}{{O{B^2}}} = \dfrac{1}{{3{a^2}}} + \dfrac{1}{{{a^2}}} = \dfrac{4}{{3{a^2}}}\) \( \Rightarrow OH = \dfrac{{\sqrt 3 a}}{2}\)

\(\Delta SOH\) vuông tại O, \(OK \bot SH \Rightarrow \dfrac{1}{{O{K^2}}} = \dfrac{1}{{O{S^2}}} + \dfrac{1}{{O{H^2}}} \Leftrightarrow \dfrac{1}{{\dfrac{{3{a^2}}}{{16}}}} = \dfrac{1}{{O{S^2}}} + \dfrac{1}{{\dfrac{3}{4}{a^2}}}\)\( \Leftrightarrow \dfrac{1}{{O{S^2}}} = \dfrac{4}{{{a^2}}} \Rightarrow SO = \dfrac{1}{2}a\)

Diện tích hình thoi ABCD: \({S_{ABCD}} = \dfrac{1}{2}AC.BD = \dfrac{1}{2}.2\sqrt 3 a.2a = 2\sqrt 3 {a^2}\)

Thể tích của khối chóp  S.ABCD là: \({V_{S.ABCD}} = \dfrac{1}{3}{S_{ABCD}}.SO = \dfrac{1}{3}.2\sqrt 3 {a^2}.\dfrac{1}{2}a = \dfrac{{\sqrt 3 {a^3}}}{3}\).

Chọn: B