Circular Motion Watch
As far as I can see, you have correctly determined that F (the force needed to maintain the particle in horizontal circular motion) = mg Newtons. Now you need to reason that the only possible thing that can be providing that force is the disc, and that the force must be in the form of friction. Next, the particle will slip if F is greater than (mu)R, where R is the vertical reaction force exerted by the disc on the particle. Can you pick it up from there?
Horizontally, there is only one force acting on the particle, namely the centripetal force acting towards the centre of rotation. It just happens that in the situation depicted by the question, the centripetal force is frictional in nature. The catch is that if the disc spins too fast, or if the value of mu is too low, the disc will be unable to provide a sufficient frictional force to maintain the particle in horizontal circular motion, and the particle will then slip outwards instead.