The Lineweaver Burk Plot is mainly interpreted by its X and Y intercepts. The question states that giving B6 increases the activity to normal levels, which means the activity should be the same as the normal. Hence, their Y intercept should be the same, so we are between choices B and C. A and D do not have the same Y intercept as normal. Higher concentrations of B6 caused activity to become normal, and so Vmax will not be changing and we can cancel A and D. A has a lower Vmax and D has a much higher Vmax. The difference between B and C is in their Km. Moving to the left on the X axis makes the Km lower and tells you that affinity is higher so you would not need more B6. But in our case affinity of the enzyme for B6 is really low, which is why we need a ton more B6.
In summary, we want the same Vmax and a higher Km. We want the "normal" activity (same vmax as normal) and we need higher amounts of B6 for success so affinity of the enzyme for B6 is probably very low.
Choice B has a (-1/Km) value closer to 0 which means Km is lower and affinity of the enzyme for its substrate is super low. This makes sense as giving higher amounts of the "competitive" substrate B6 is helping.
Most problems with cell division occur in anaphase of Meiosis I.