Chọn D

$\overrightarrow{C{A}^{\text{'}}}=\overrightarrow{CD}+\overrightarrow{CB}+\overrightarrow{C{C}^{\text{'}}}\Rightarrow {\overrightarrow{C{A}^{\text{'}}}}^2={\left(\overrightarrow{CD}+\overrightarrow{CB}+\overrightarrow{C{C}^{\text{'}}}\right)}^2\iff 8a^2=4a^2+a^2+5a^2+2.a.a\sqrt{5}\cos C^{\text{'}}CB$

Suy ra $\cos C^{\text{'}}CB=-\frac{1}{\sqrt{5}}=\cos D^{\text{'}}DA\Rightarrow \cos A^{\text{'}}AD=\frac{1}{\sqrt{5}}$.

Gọi H là hình chiếu của A' trên AD. Có $\cos A^{\text{'}}AD=\frac{AH}{A{A}^{\text{'}}}=\frac{1}{\sqrt{5}}\Rightarrow AH=a$ nên $H\equiv D$.

Suy ra $A{D}^{\text{'}}=2a; S_{AD{D}^{\text{'}}{A}^{\text{'}}}=A^{\text{'}}D.DA=2a^2$.

Có $CD\perp DA;CD\perp D{D}^{\text{'}}\Rightarrow CD\perp \left(AD{D}^{\text{'}}{A}^{\text{'}}\right)\Rightarrow V_{ABCD.A^{\text{'}}{B}^{\text{'}}{C}^{\text{'}}{D}^{\text{'}}}=CD.S_{AD{D}^{\text{'}}{A}^{\text{'}}}=4a^3$.