You are overthinking this. If v already happens to lie in that given subspace, then Pv= v. What eigenvalue does that correspond to? And, of course, the "multiplicity" is the dimension of that subspace. If v is orthogonal to that subspace, then Pv= 0. What eigenvalue does that correspond to. The "multiplicity" of that eigenvalue is the dimension of V minus the dimension of the given subspace.
By the way, you "don't know how to find the eigenvalues of a matrix that isn't square" because non-square matrices don't have eigenvalues! is an eigenvalue of a matrix (or, more generally, linear transformation), A, if and only for some non-zero eigenvector.
In order for that to even make sense, v must be a member of both the "domain" and "range" spaces for A. That is, A must map some vector space V into itself which means that the matrix representing A must be square.