 Solution

Put $a={\lambda }_P$ and $b={\lambda }_Q$. The Pauli operators are Hermitian and satisfy $P^2=Q^2=I$ and $PQ+QP=0$, soThus $V$ is a unitary operator. For every Pauli operator $R$, according as $R$ commutes or anticommutes with $P$ and $Q$, expansion of $VRV$ gives one of $\pm R$ or $\pm RPQ$, up to the phase that makes it Hermitian. It is therefore another Pauli operator, so $V$ normalizes the Pauli group and is a Clifford operation. Finally,which is the normalized projection onto the eigenvalue-$b$ eigenspace of $Q$. Hence $V$ maps the eigenvalue-${\lambda }_P$ eigenspace of $P$ onto the eigenvalue-${\lambda }_Q$ eigenspace of $Q$.