C

Step 1: First, analyze the forces acting on the mass M. Since there are two tension forces due to the strings making an angle \( \theta \) with the vertical, we can denote the tension in each string as T.

Step 2: The vertical component of the tension forces helps in balancing the downward acceleration 'a' and the weight of the mass M. The vertical component of the tension can be expressed as \( T \cos \theta \). Therefore, for both sides, we can express the equation as:
\[ 2T \cos \theta = M(g - a) \]
where 'g' is the acceleration due to gravity.

Step 3: We need to express T in terms of acceleration 'a'. Rearranging gives:
\[ T = \frac{M(g - a)}{2 \cos \theta} \]
This indicates that to achieve the downward acceleration equal to 'a', the tension has to compensate for the change from gravity, which leads to a downward acceleration that modifies the effective gravitational pull on mass M.

Step 4: The mass M will accelerate at an effective acceleration downward given the tension and angle, simplifying down. We find that replacing this into our earlier derived expression leads us to:
\[ a_{effective} = a \cdot \sec \theta. \]

Therefore, the answer that corresponds to this effective acceleration is:
\( a_{effective} = a \sec \theta \), hence the correct option is: C.