Recently I had gone through a statistical physics course and I learned about phase space. A point in phase space representing the state of the system of $``N"$ particles. $``N"$ can be any number. And if we have the phase space distribution function, say $f(p,q)$, $p$ is the canonical momentum and $q$ is the generalized coordinate, then $$f(p,q)d\tau$$ gives the probability of finding the system in that particular state in the $dp$$dq$ range. $$\int f(p,q)d\tau=1$$ But now after reading Landau's Physical Kinetics which is Vol. $10$. It is stating otherwise. Why is there two other definition of distribution functions in phase space? The doubt is in the product of distribution function and the volume element, sometimes it says the mean number of molecules and sometimes it says the probability. How is this possible with phase space distributions?
These are the same definitions. I think there is a misunderstanding in the use of the language when you say "product of distribution function and the volume element"
The first equation you have written defines the phase space to present it in 2D, based upon the coordinates and the momentum, for the N particles. The second equation is the statement that you have a normalized probability distribution over that phase space so that the densities are actual probabilities that do not need re-normalization; this phase space exhausted by the integral over $d\tau$. In the passage you cite the 'p' and 'q' are defined to represent the same quantities. Since $f(p,q)d\tau$ ($f d\tau$ in the passage) is a continuous distribution, we need to refer to expected value (mean value as stated in that passage since I assume uniformity in the phase space) for a region. This product you see is the need to get a value from the continuous probability distribution into a probability for the space. Over the trajectory of the phase space, some molecules will be in the state f(t,p,q) and some in f(t',p',q') and the probabilities will not be the same.