Gọi I là trung điểm AD \(\Rightarrow SI\perp AD\Rightarrow SI\left(ABCD\right)\Rightarrow d\left(I;\left(ABCD\right)\right)=SI\)
Ta có \(SM\cap\left(ABCD\right)=\left\{B\right\}\) và \(\frac{SB}{MB}=2\) nên \(d\left(M;\left(ABCD\right)\right)=\frac{1}{2}d\left(I;\left(ABCD\right)\right)=\frac{1}{2}SI=\frac{1}{2}\cdot\frac{a\sqrt{3}}{2}=\frac{a\sqrt{3}}{4}\)
\(S_{CNP}=\frac{1}{2}\cdot CN\cdot CP=\frac{1}{2}\cdot\frac{1}{2}CD\cdot\frac{1}{2}\cdot BC=\frac{a^2}{8}\)
\(V_{M.CNP}=\frac{1}{3}\cdot d\left(M;\left(ABCD\right)\right)\cdot S_{CNP}=\frac{a^3\sqrt{3}}{96}\)