Similar matrices have the same eigenvalues

Characteristic polynomial. Since $P^{-1}(XI)P=XI$,
$$\det(XI-B)=\det\big(P^{-1}(XI-A)P\big)=\det(P^{-1})\det(XI-A)\det(P)=\det(XI-A),$$
because $\det(P^{-1})\det(P)=\det I=1$.

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

Eigenspaces. If $Av=\lambda v$, then $B(P^{-1}v)=P^{-1}APP^{-1}v=P^{-1}Av=\lambda P^{-1}v$. So $v\mapsto P^{-1}v$ maps $E_\lambda(A)$ into $E_\lambda(B)$, and it is injective, so $\dim E_\lambda(A)\le\dim E_\lambda(B)$. Since $A=PBP^{-1}$, the same argument with $P$ gives the reverse inequality. Hence $\dim E_\lambda(A)=\dim E_\lambda(B)$.

Conclusion. Similar matrices have identical spectral data (eigenvalues, algebraic and geometric multiplicities).