Answer : A - Regular
$L = \left \{ a^ib^jc^kd^m | i + j+k+m \,\,is\,\,multiple\,\,of\,\,13 \right \}$ is nothing but an Intersection of Two Regular languages.
$L = L_1 \cap L_2 = \left \{ a^*b^*c^*d^* \right \} \bigcap \left \{ w | w \in \left \{ a+b+c+d \right \}^*, Length \,\,of\,\, w \,\,is \,\,divisible\,\, by \,\,13 \right \}$
Both $L_1$ = $\left \{ a^*b^*c^*d^* \right \} $ and $L_2 = $ $ \left \{ w | w \in \left \{ a+b+c+d \right \}^*, Length \,\,of\,\, w \,\,is \,\,divisible\,\, by \,\,13 \right \}$
are Regular as First one($L_1$) is a Regular expression itself and the Second one($L_2$) is a Mod 13 language.
NOTE : Since $L$ is Regular, Hence, All the Options are Correct. But the Strongest answer would be Regular.