Since $M_1$ and $M_4$ are non singular, both $M_1^{-1}$ and $M_4^{-1}$ exist.
Given
$M_1M_2M_3M_4=0$,
multiply on the left by $M_1^{-1}$ and on the right by $M_4^{-1}$. Then we get
$M_1^{-1}(M_1M_2M_3M_4)M_4^{-1}=M_1^{-1}\,0\,M_4^{-1}$,
which gives
$M_2M_3=0$.
So option $\boxed{\text{A}}$ is true.
Now check the other options.
Option B is not necessary. Two nonzero matrices can have zero product.
For example, take
$M_2=\begin{pmatrix}1&0\\0&0\end{pmatrix}$, $\qquad M_3=\begin{pmatrix}0&0\\0&1\end{pmatrix}$.
Then $M_2\neq 0$, $M_3\neq 0$, but
$M_2M_3=0$.
So B is false.
Option C is false because $M_2M_3=0$, and the zero matrix is never non singular.
Option D is also false because nothing in the given condition implies $M_1M_4=0$.
Hence, the correct answer is A.
