The problem seems poorly worded; the intended meaning—the only possible self-consistent description—appears to be that the particle comes to rest with respect to the disc after impact.
(Comments on the page discuss at least two other possibilities that can't occur, either by the problem description or Nature's restrictions. The first is that the particle rebounds elastically from the disc and immediately gets stuck on the table to end up at rest. But a frictionless surface provides no such opportunity for sticking. The second is that kinetic energy is gained by the disc and the system cools down. But since cooling a system decreases its entropy and the work done to accelerate something transfers no entropy, entropy must be destroyed, which Nature prohibits.)
The initial momentum (of the particle) is $mv$. Linear momentum is conserved for the objects on a frictionless table, so the final speed of the mass of $2m$ is $\frac{(mv)}{2m}=\frac{v}{2}$.
The initial angular momentum around the disc center (of the particle) is $(mv)\frac{R}{2}$. Angular momentum is conserved as well for the objects on a frictionless table, so the final rotational speed is $(mv)\frac{R}{2}\left(\frac{1}{I}\right)$, where $I$ is the mass moment of inertia for the particle and disc rotating in concert around their common center of mass, which is $R/2$ away from the center. (Thank you to
@Albertus Magnus for identifying this behavior.)
From the parallel axis theorem, $I=I_\mathrm{circle}+I_\mathrm{particle}=\left(\frac{mR^2}{2}+\frac{mR^2}{4}\right)+\frac{mR^2}{4}=mR^2$. This angular speed is thus $\frac{v}{2R}$.
The initial kinetic energy is $\frac{1}{2}mv^2$. The final kinetic energy is $\frac{1}{2}(2m)\left(\frac{v}{2}\right)^2+\frac{1}{2}\left(mR^2\right)\left(\frac{v}{2R}\right)^2=\frac{3}{8}mv^2$, which is lower after this inelastic collision.