Can the angular momentum of Rod be conserved about point B although the point B is accelerated during collision?
If you mean B as a point in an inertial frame, then yes. If you mean B as a point attached to the rod, then no. You're correct that if the axis under consideration is accelerating then it makes things difficult. In general, you don't want to do this unless the accelerating axis contains the center of mass of the object you are considering to rotate.
Does elastic collision mean that the Velocity of approach = velocity of seperation about Common Normal through point B?
It means that the total mechanical energy of the system (all kinetic in this case) is the same before and after the collision. In a frame where the center of mass is at rest, then the velocities before and after will be the same. In other frames, it may not be. A golf ball and a golf club may have a nearly elastic collision, but the ball will have a very different velocity before and after being struck on a tee.
This setup needs to be thought of in three distinct phases: before the collision, a small time $\Delta t$ around the moment of the collision, and finally the motion after the collision while the ball is rising. Your question is not 100% clear about which phase you're concerned with, so I'll give a brief description of all three.
Before the ball hits the "pointy edge", gravity does indeed exert a torque on the ball if we take the "pointy edge" to be our origin point. So does the normal force from the surface it is rolling on, before it hits the edge. Since these two forces are equal and opposite and act along the same line, they exert opposite torques before the ball hits the pointy edge, and so angular momentum is constant with respect to this point.
In a small amount of time $\Delta t$ around the moment of the collision, the ball may experience an impulsive force from the pointy edge; this exerts no torque. In addition, in this brief amount of time, the torque from gravity will be approximately $m g \ell$, where $\ell$ is the horizontal distance between the edge and the center of the ball. This means that the change in angular momentum of the ball in this time will be approximately $m g \ell \Delta t$, which is negligible if we take the limit $\Delta t \to 0$. Thus, in the collision with the edge, angular momentum is conserved. This means that the angular momentum of the ball immediately after the collision is the same as it was long before the collision. (It is not immediately clear to me whether energy is conserved in the time $\Delta t$ around the collision; I suspect it is not.)
Finally, once the ball has lost contact with the surface below, the normal force vanishes, while gravity still exerts a torque. As before, the contact force exerts no torque on the ball. Thus, there is now a net torque on the ball, and the angular momentum with respect to the pivot point will no longer be constant. This means that we can't use angular momentum conservation to make statements about the motion of the ball while it is "tripping" over the edge. Energy is still conserved from the moment after the collision onwards, though. (This is important for some problems you might ask about this setup.)
Best Answer
At the very moment that the bullet sticks to the rod, the rod remains vertical and therefore the weight of the bullet is downward along the line passing through the hinge. The torque then is zero. You can, therefore, apply the conservation of momentum to the moment just before and just after the collision.
Following the collision, the rod begins to rise due to the impulse transferred to it and as it is no longer vertical, the torque due to its weight will no longer be zero. Therefore you cannot apply the conservation of angular momentum for the motion following the collision. (But you can apply conservation of energy now!)