A particle is projected from point O with velocity u in a direction making an angle α with the horizontal. At any instant its position is at point P at right angles to the initial direction of projection. Its velocity at point P is

The correct option is u cotα.

Let us analyze the motion of the particle. The particle is projected from the origin $O$ with an initial velocity vector $\overrightarrow{u}$ at an angle $\alpha$ to the horizontal.
We can express the initial velocity vector in Cartesian coordinates as:
$\overrightarrow{u}=u\cos \alpha \hat{i}+u\sin \alpha \hat{j}$

At any subsequent time $t$, let the velocity of the particle at point $P$ be $\overrightarrow{v}$.
In projectile motion, there is no acceleration in the horizontal direction. Therefore, the horizontal component of velocity remains constant throughout the motion:
$v_x=u\cos \alpha$

Let the velocity vector at $P$ be:
$\overrightarrow{v}=v_x\hat{i}+v_y\hat{j}=u\cos \alpha \hat{i}+v_y\hat{j}$

We are given that at point $P$, the direction of motion (which is along the velocity vector $\overrightarrow{v}$) is at right angles to the initial direction of projection $\overrightarrow{u}$.
Since the two vectors $\overrightarrow{u}$ and $\overrightarrow{v}$ are perpendicular, their dot product must be zero:
$\overrightarrow{u}·\overrightarrow{v}=0$

Substituting the component forms:
$(u\cos \alpha \hat{i}+u\sin \alpha \hat{j})·(u\cos \alpha \hat{i}+v_y\hat{j})=0$
$u^2{\cos }^2\alpha +u{v}_y\sin \alpha =0$

Solving for $v_y$:
$u{v}_y\sin \alpha =-u^2{\cos }^2\alpha$
$v_y=-\frac{u{\cos }^2\alpha }{\sin \alpha }$

Now, the magnitude of the velocity at point $P$ is given by:
$v=\sqrt{{v_x}^2+{v_y}^2}$

Substituting $v_x$ and $v_y$:
$v=\sqrt{(u\cos \alpha {)}^2+{(-\frac{u{\cos }^2\alpha }{\sin \alpha })}^2}$
$v=\sqrt{u^2{\cos }^2\alpha +\frac{u^2{\cos }^4\alpha }{{\sin }^2\alpha }}$

Factoring out $u^2{\cos }^2\alpha$ from inside the square root:
$v=u\cos \alpha \sqrt{1+\frac{{\cos }^2\alpha }{{\sin }^2\alpha }}$
$v=u\cos \alpha \sqrt{\frac{{\sin }^2\alpha +{\cos }^2\alpha }{{\sin }^2\alpha }}$

Since ${\sin }^2\alpha +{\cos }^2\alpha =1$, we have:
$v=u\cos \alpha \sqrt{\frac{1}{{\sin }^2\alpha }}$
$v=u\frac{\cos \alpha }{\sin \alpha }$
$v=u\cot \alpha$

Thus, the magnitude of the velocity of the particle at point $P$ is $u\cot \alpha$.