$SB=\sqrt{(2a)^2+a^2}=a\sqrt5$
Ta có: $\sin(SB,(SCD))=\dfrac{d (B; (SCD))}{SB}$
Mà $AB//(SCD)$ vì $AB//CD$
$\to d(B; (SCD))=d(A;(SCD))$
Kẻ $AI\bot SD$
Ta có $SA\bot CD, CD\bot AD$
Suy ra $CD\bot(SAD)$
$\to CD\bot AI$
Mà $AI\bot SD$ nên $AI\bot(SCD)$
$\to d(A;(SCD))=AI$
$\Delta SAD$ vuông tại $A$, đường cao $AI$ có:
$\dfrac{1}{AI^2}=\dfrac{1}{SA^2}+\dfrac{1}{AD^2}=\dfrac{1}{4a^2}+\dfrac{1}{a^2}$
$\to AI=\dfrac{2a}{\sqrt5}=d(B;(SCD))$
$\to \sin(SB,(SCD))=\dfrac{ \dfrac{2a}{\sqrt5} }{ a\sqrt5}=\dfrac{2}{5}$