Let V be the speed of bead B
Now since the string remains tight velocity component of A and B along the string AB must be same
$u \cos \theta=v \cos \alpha$
$\theta=45^{\circ}$ and by geometring we get
$\alpha=45^{\circ}$
$u \large\frac{1}{\sqrt 2}$$=v \times \large\frac{1}{\sqrt2}$
$v=\ u$
Hence b is the correct answer.