Stationary particle (quantum) Setting the potential oprator to zero in the Schrodinger Equation gives us the 
description of a free particle. Will setting the momentum operator (squared) to zero gives us a meaningful description of a stationary particle?  
 A: This is probably not what the OP had in mind but it is common practice not to include the $p^2/2m$ term to study spins on a lattice for example, where one is only interested in the spin degrees of freedom, completely neglecting the positional ones. Look for yourself at the Hamiltonian of the Ising Model here: no momentum operator in it. 
As pointed out in a comment to this answer, this is an approximation, as a complete Hamiltonian would feature the momentum of the particles carrying the spin. What happens here, usually, is that the positional degrees of freedom decouple from the spin ones, and it is therefore legitimate to consider a Hamiltonian just for the spins. In the context of such a lattice, i.e. physically a crystal, the positional degrees of freedom would typically give rise to quasi-particle called phonons, which are the quantum of the lattice vibration, and a complete Hamiltonian would then induce an interaction between the phonons and the spins: I don't mean a direct coupling term in the Hamiltonian but that the spin-spin coupling would depend on the distance between the particle.
A: You can't just set the momentum operator to zero. Momentum is an intrinsic property of a particle and the momentum operator measures it. The potential is determined by you, so if you don't supply one its operator may vanish.
What you can do, however, is to look for a state with zero momentum
$$\hat{p}\psi=0$$
which has a solution
$$\psi=\frac{1}{\sqrt{2\pi}}$$
with Dirac normalization.
In summary, in the case you described you don't set the operator to zero - you set its action on the wavefunction to zero.
