Let v be an eigen vector of A, such that v*A = sv for some scaling factor s. Take the preimage of v under P. In other words, w*P = v. Now w*P*A = sv, and when we apply P inverse, we get sw. Thus w is an eigen vector for B, with the same eigen value s.
By symmetry, an eigen vector w of the matrix B becomes an eigen vector v of the matrix A, with the same eigen value s.
If A and B are similar, eigen values and eigen spaces correspond. Use P or P inverse to map one eigen space to the other.
This makes sense if you think geometrically. If A has an eigen vector v, the linear transformation implemented by P is going to move some vector w onto v. Once v is scaled by A, the inverse of P will pull it back to a scaled version of w. Thus B has the same eigen vectors as A, they're just moved around a bit.
All this shows the set of eigen values is the same, but doesn't verify multiplicity.
We can show equal characteristic polynomials, and hence the same eigen values,
by looking at B-s*I, where s is a variable that represents the eigen value, and I is the identity matrix.
This is the same as P*A/P-s*I,
or P*A/P-P*s*I/P,
or P*(A-s*I)/P.
Take determinants, and the determinant of B, with s subtracted from the main diagonal, is the same as the determinant of A, with s subtracted from the main diagonal. Both matrices produce the same characteristic polynomial, and both matrices have the same eigen values, multiplicities included.