Ta có: \(SA \bot \left( {ABCD} \right) \Rightarrow SA \bot CD\)

Lại có: \(CD \bot AB\;\left( {gt} \right) \Rightarrow CD \bot \left( {SAB} \right) \Rightarrow CD \bot SD \Rightarrow \Delta SDC\) vuông tại \(D.\)

Có \(\left( {ABCD} \right) \cap \left( {SCD} \right) = CD.\)

Mà:\(\left\{ \begin{array}{l}CD \bot SD\;\;\left( {cmt} \right)\\CD \bot AD\;\;\left( {gt} \right)\end{array} \right. \Rightarrow \left( {\left( {ABCD} \right),\;\left( {SCD} \right)} \right) = \left( {SD,\;AD} \right) = \angle SDA = {60^0}\)

Xét \(\Delta SAD\) vuông tại \(A\) ta có: \(SA = AD.\tan {60^0} = AD\sqrt 3 .\)

\(\begin{array}{l}{V_{SABCD}} = \dfrac{{{a^3}\sqrt 3 }}{3} \Leftrightarrow \dfrac{1}{3}A{D^2}.AD\sqrt 3  = \dfrac{{{a^3}\sqrt 3 }}{3} \Leftrightarrow AD = a.\\ \Rightarrow {V_{SACD}} = \dfrac{1}{2}{V_{SACD}} = \dfrac{1}{2}.\dfrac{{{a^3}\sqrt 3 }}{3} = \dfrac{{{a^3}\sqrt 3 }}{6} = \dfrac{1}{3}d\left( {A;\;\left( {SCD} \right)} \right).{S_{SCD}}.\\ \Rightarrow d\left( {A;\;\left( {SCD} \right)} \right) = \dfrac{{3{V_{SACD}}}}{{{S_{SCD}}}}.\end{array}\)

Ta có: \(SD = \sqrt {S{A^2} + A{D^2}}  = \sqrt {3{a^2} + {a^2}}  = 2a.\)

\(\begin{array}{l} \Rightarrow {S_{SCD}} = \dfrac{1}{2}SD.CD = \dfrac{1}{2}.2a.a = {a^2}.\\ \Rightarrow d\left( {A;\;\left( {SCD} \right)} \right) = \dfrac{{3{V_{SACD}}}}{{{S_{SCD}}}} = \dfrac{{3.\dfrac{{{a^3}\sqrt 3 }}{6}}}{{{a^2}}} = \dfrac{{a\sqrt 3 }}{2}.\end{array}\)

Vì \(AB//CD \Rightarrow AB//\left( {SCD} \right) \Rightarrow d\left( {A;\;\left( {SCD} \right)} \right) = d\left( {B;\;\left( {SCD} \right)} \right)\)

Lại có: \(\dfrac{{MC}}{{BC}} = \dfrac{1}{2}\left( {gt} \right) \Rightarrow \dfrac{{d\left( {M;\;\left( {SCD} \right)} \right)}}{{d\left( {B;\;\left( {SCD} \right)} \right)}} = \dfrac{1}{2}\)

\( \Rightarrow d\left( {M;\;\left( {SCD} \right)} \right) = \dfrac{1}{2}d\left( {B;\;\left( {SCD} \right)} \right) = \dfrac{1}{2}d\left( {A;\;\left( {SCD} \right)} \right) = \dfrac{1}{2}.\dfrac{{a\sqrt 3 }}{2} = \dfrac{{a\sqrt 3 }}{4}.\)

Chọn C.