Chọn phương án A.

Gọi \(G_1,\,G_2,\,G_3,\,G_4\) lần lượt là trọng tâm \(\triangle SAB\), \(\triangle SBC\), \(\triangle SCD\), \(\triangle SDA\).

Gọi \(E,\,F,\,G,\,H\) lần lượt là trung điểm của các cạnh \(AB\), \(BC\), \(CD\), \(DA\).

• Vì \(\dfrac{OG_1}{OM}=\dfrac{OG_2}{ON}=\dfrac{OG_3}{OP}=\dfrac{OG_4}{OQ}=\dfrac{1}{2}\) nên \(\dfrac{S_{G_1G_2G_3G_4}}{S_{MNPQ}}=\dfrac{1}{4}\)
• Vì \(\dfrac{SG_1}{SE}=\dfrac{SG_2}{SF}=\dfrac{SG_3}{SG}=\dfrac{SG_4}{SH}=\dfrac{2}{3}\) nên \(\dfrac{S_{G_1G_2G_3G_4}}{S_{MNPQ}}=\dfrac{4}{9}\)
• Vì \(ABCD\) là hình vuông nên \(EFGH\) cũng là hình vuông.
  Suy ra \(S_{EFGH}=EF^2=\left(\dfrac{a\sqrt{2}}{2}\right)^2=\dfrac{a^2}{2}\).

Ta có $$\begin{aligned}S_{MNPQ}&=4S_{G_1G_2G_3G_4}=4\cdot\dfrac{4}{9}{S_{EFGH}}\\ &=4\cdot\dfrac{4}{9}.\dfrac{a^2}{2}=\dfrac{8a^2}{9}\end{aligned}$$
Vì \(S.ABCD\) là hình chóp đều nên \(SO\bot(ABCD)\).
Suy ra \(SS'\bot(ABCD)\). Do đó:$$\begin{aligned}
\mathrm{d}\left(S',\left(MNPQ\right)\right)&=\mathrm{d}\left(S',\left(ABCD\right)\right)+\mathrm{d}\left(O,\left(MNPQ\right)\right)\\
&=\mathrm{d}\left(S,\left(ABCD\right)\right)+2\mathrm{d}\left(O,\left(G_1G_2G_3G_4\right)\right)\\
&=\mathrm{d}\left(S,\left(ABCD\right)\right)+2\cdot\dfrac{1}{3}\mathrm{d}\left(S,\left(ABCD\right)\right)\\
&=\dfrac{5}{3}\mathrm{d}\left(S,\left(ABCD\right)\right)=\dfrac{5}{3}SO\\
&=\dfrac{5}{3}\sqrt{SA^2-OA^2}=\dfrac{2}{3}\sqrt{(2a)^2-\left(\dfrac{a\sqrt{2}}{2}\right)^2}\\
&=\dfrac{5a\sqrt{14}}{6}.
\end{aligned}$$
Vậy \(V_{S'.MNPQ}=\dfrac{1}{3}\cdot\dfrac{5a\sqrt{14}}{6}\cdot\dfrac{8a^2}{9}=\dfrac{20a^3\sqrt{14}}{81}\).