$F = qvB$ (Force of charged particle in a magnetic field)
And we know that
$F =\frac{ mv ^2}{ r } \quad$ (r is the radius of motion and $m$ is mass of particle)
$\Rightarrow qvB =\frac{ mv ^2}{ r }$
$\Rightarrow r =\frac{ mv }{ Bq }$
Now as we know that
$\omega=\frac{ V }{ r }$
$\Rightarrow \omega=\frac{ Bq }{ m }$
Time period, $T =\frac{2 \pi}{\omega}$
$\Rightarrow T =\frac{2 \pi m }{ Bq }$
And this shows that it is independent of both radius and velocity.

$[A$ For $B>\frac{2}{3} \frac{p}{QR}$, the particle will re-enter region $1$
$[B]$ For $B=\frac{8}{13} \frac{\mathrm{p}}{QR}$, the particle will enter region $3$ through the point $P_2$ on $\mathrm{x}$-axis
$[C]$ When the particle re-enters region 1 through the longest possible path in region $2$ , the magnitude of the change in its linear momentum between point $P_1$ and the farthest point from $y$-axis is $p / \sqrt{2}$
$[D]$ For a fixed $B$, particles of same charge $Q$ and same velocity $v$, the distance between the point $P_1$ and the point of re-entry into region $1$ is inversely proportional to the mass of the particle