Similar matrices have the same eigenvalues

Characteristic polynomial.
$$\det(B-\lambda I)=\det(P^{-1}AP-\lambda I)=\det(P^{-1}(A-\lambda I)P)=\det(P^{-1})\det(A-\lambda I)\det(P).$$
Since $\det(P^{-1})\det(P)=1$, we get $\det(B-\lambda I)=\det(A-\lambda I)$.

Consequence. $\chi_A=\chi_B$, so $A$ and $B$ have the same eigenvalues with the same multiplicities.

Eigenvectors. If $Av=\lambda v$, then $B(P^{-1}v)=P^{-1}Av=\lambda P^{-1}v$. So $P^{-1}$ maps eigenspaces of $A$ to eigenspaces of $B$. In particular, $\dim E_\lambda(A)=\dim E_\lambda(B)$.

Conclusion. Similar matrices have identical spectral data.