How to choose a proper reference point to apply conservation of angular momentum?

I am having trouble in applying the conservation of angular momentum in the following problem.

A uniform rod of mass M and length a lies on a smooth horizontal plane. A particle of
mass m moving at a speed v perpendicular to the length of the rod strikes it at a
distance a/4 from the center and stops after the collision. the angular velocity of the rod about its center just after the collision.

My answer when solved from the point of reference of A does not match to that of center of mass

while taking A as a reference point the following points can be deduced-

Initial conditions:

$$ L _{rod} = 0$$

$$L _{particle} = frac{mva}{4}$$

Final condition:

$$ L _{particle} = 0$$

$$L _{rod} = mathbf{L _{com}} + M(mathbf{r _{o} x v })$$

because ro and v are in same direction.

Hence,

$$mathbf{r _{o} x v } = 0$$

and

$$L_{com} = I omega $$

So,

$$L _{rod} = I omega $$

Now applying conservation of angular momentum we get

$$frac{mva}{4} = frac{Ma^{2} omega }{12} $$

Now we can solve for angular velocity.

If we move over to com of rod and try to solve for angular momentum.

We know that from the reference of point A, rod starts to move translationally and particle comes to rest. Translational speed of rod can be calculated using conservation of linear momentum which is mv/M

Now initial conditions to apply the conservation of angular momentum for point A and com will be same. For final conditions for com:

$$L _{rod} = I omega $$

Because center of mass is moving with velocity mv/M so speed of particle seen from center of mass will be -mv/M

$$L _{particle} = frac{-m ^{2} av }{4M}$$

Writing the equation of conservation of angular momentum it will give:

$$frac{mva}{4}= Iomega – frac{m ^{2} av }{4M}$$

Answers for angular velocity from both the equations does not match what am I doing wrong.