Gắn hệ trục tọa độ như hình vẽ, giả sử \(ABCD\) là hình vuông cạnh \(1\), chiều cao hình chóp \(SH = h\).

Khi đó \(A\left( {0;0;0} \right),B\left( {1;1;0} \right),D\left( {0;1;0} \right),C\left( {1;1;0} \right)\).

Gọi tọa độ \(H\left( {a;b;0} \right) \Rightarrow S\left( {a;b;h} \right)\).

Ta có: \(\overrightarrow {AS}  = \left( {a;b;h} \right),\overrightarrow {AD}  = \left( {0;1;0} \right) \Rightarrow {\overrightarrow n _{\left( {SAD} \right)}} = \left[ {\overrightarrow {AS} ,\overrightarrow {AD} } \right] = \left( { - h;0;a} \right)\)

\(\overrightarrow {BS}  = \left( {a - 1;b;c} \right),\overrightarrow {BC}  = \left( {0;1;0} \right)\) \( \Rightarrow {\overrightarrow n _{\left( {SBC} \right)}} = \left[ {\overrightarrow {BS} ,\overrightarrow {BC} } \right] = \left( { - h;0;a - 1} \right)\)

\(\overrightarrow {AB}  = \left( {1;0;0} \right),\overrightarrow {AS}  = \left( {a;b;h} \right) \Rightarrow {\overrightarrow n _{\left( {SAB} \right)}} = \left[ {\overrightarrow {AB} ,\overrightarrow {AS} } \right] = \left( {0; - h;b} \right)\)

\({\overrightarrow n _{\left( {ABCD} \right)}} = \overrightarrow k  = \left( {0;0;1} \right)\).

Do \(\left( {SAD} \right) \bot \left( {SBC} \right) \Rightarrow {\overrightarrow n _{\left( {SAD} \right)}}.{\overrightarrow n _{\left( {SBC} \right)}} = 0 \Leftrightarrow {h^2} + a\left( {a - 1} \right) = 0 \Leftrightarrow {h^2} + {a^2} = a\,\,\left( 1 \right)\)

Góc giữa \(\left( {SAB} \right)\) và \(\left( {SBC} \right)\) bằng \({60^0}\) \( \Rightarrow \cos {60^0} = \dfrac{{\left| {{{\overrightarrow n }_{\left( {SAB} \right)}}.{{\overrightarrow n }_{\left( {SBC} \right)}}} \right|}}{{\left| {{{\overrightarrow n }_{\left( {SAB} \right)}}} \right|.\left| {{{\overrightarrow n }_{\left( {SBC} \right)}}} \right|}} \Leftrightarrow \dfrac{1}{2} = \dfrac{{\left| {b\left( {a - 1} \right)} \right|}}{{\sqrt {{h^2} + {{\left( {a - 1} \right)}^2}} .\sqrt {{h^2} + {b^2}} }}\)

\( \Leftrightarrow \dfrac{1}{2} = \dfrac{{b\left( {a - 1} \right)}}{{\sqrt {1 - a} \sqrt {{h^2} + {b^2}} }} \Leftrightarrow \dfrac{{b\sqrt {1 - a} }}{{\sqrt {{h^2} + {b^2}} }} = \dfrac{1}{2} \Leftrightarrow \dfrac{b}{{\sqrt {{h^2} + {b^2}} }} = \dfrac{1}{{2\sqrt {1 - a} }}\)

Góc giữa \(\left( {SAB} \right)\) và \(\left( {SAD} \right)\) là \({45^0}\) \( \Rightarrow \cos {45^0} = \dfrac{{\left| {{{\overrightarrow n }_{\left( {SAB} \right)}}.{{\overrightarrow n }_{\left( {SAD} \right)}}} \right|}}{{\left| {{{\overrightarrow n }_{\left( {SAB} \right)}}} \right|.\left| {{{\overrightarrow n }_{\left( {SAD} \right)}}} \right|}} \Leftrightarrow \dfrac{{\sqrt 2 }}{2} = \dfrac{{\left| {ab} \right|}}{{\sqrt {{h^2} + {a^2}} \sqrt {{h^2} + {b^2}} }}\) \( \Leftrightarrow \dfrac{{\sqrt 2 }}{2} = \dfrac{{ab}}{{\sqrt a .\sqrt {{h^2} + {b^2}} }}\)

Suy ra \(\dfrac{{ab}}{{\sqrt a .\sqrt {{h^2} + {b^2}} }}:\dfrac{{b\sqrt {1 - a} }}{{\sqrt {{h^2} + {b^2}} }} = \dfrac{{\sqrt 2 }}{2}:\dfrac{1}{2} \Leftrightarrow \dfrac{{\sqrt a }}{{\sqrt {1 - a} }} = \sqrt 2  \Leftrightarrow a = \dfrac{2}{3}\).

Gọi góc giữa \(\left( {SAB} \right)\) và \(\left( {ABCD} \right)\) là \(\alpha \) \( \Rightarrow \cos \alpha  = \dfrac{{\left| {{{\overrightarrow n }_{\left( {SAB} \right)}}.{{\overrightarrow n }_{\left( {ABCD} \right)}}} \right|}}{{\left| {{{\overrightarrow n }_{\left( {SAB} \right)}}} \right|\left| {{{\overrightarrow n }_{\left( {ABCD} \right)}}} \right|}} = \dfrac{b}{{\sqrt {{h^2} + {b^2}} }} = \dfrac{1}{{2\sqrt {a - \dfrac{2}{3}} }} = \dfrac{{\sqrt 3 }}{2}\).

Chọn C.