The first dependency to consider is the dependency induced by the primary key:
$M\to$ all attributes.
Since $M$ is a key, it does not violate BCNF.
Next,
$M\to D$ also does not violate BCNF because $M$ is a superkey.
Now consider
$E\to R$.
$E$ is not a superkey.
Therefore, this is the first BCNF violation.
Decompose into $R_1(E,R)$ and a remaining relation containing
$D,M,N_1,N_2,G_1,G_2,C_1,C_2,E$.
$ER$ is in BCNF because $E$ is its key.
Next consider
$N_1\to G_1C_1$.
$N_1$ is not a superkey of the large remaining relation.
Decompose out $R_2(N_1,G_1,C_1)$.
Similarly,
$N_2\to G_2C_2$ produces $R_3(N_2,G_2,C_2)$.
The remaining schema is $R_4(M,D,N_1,N_2,E)$.
Because $M$ was the primary key of the original relation, its projected dependencies still allow it to determine the attributes of this component.
Therefore, $R_4$ is in BCNF.
Now check the two reverse dependencies.
Inside $R_2(N_1,G_1,C_1)$, we have both $N_1\to G_1C_1$ and $G_1C_1\to N_1$.
Thus both $N_1$ and $G_1C_1$ are candidate keys of $R_2$.
So no BCNF violation remains.
Exactly the same reasoning applies to $R_3(N_2,G_2,C_2)$.
Hence, the final decomposition is $\boxed{\{ER,\ N_1G_1C_1,\ N_2G_2C_2,\ MDN_1N_2E\}}$
Answer : D
