It seems to me that we are here considering the parity operator $P$ simultaneously an observable (thus selfadjoint) and a symmetry (thus unitary or antiunitary from Wigner's theorem). Antiunitary and selfadjoint is impossible, so that the only possibility is that $P$ is both unitary and selfadjoint.
Every unitary operator has spectrum contained in $\{\lambda \in \mathbb{C} \:|\: |\lambda|=1\}$. Every selfadjoint operator has real spectrum.
Therefore if $P$ is simultaneously unitary and selfadjoint, its eigenvalues must be in the set $\{\pm 1\}$ which is the intersection of the sets above.
Barring trivial cases, the set of eigenvalues of $P$ must coincide with that whole set $\{\pm 1\}$ actually. In the other cases $P$ would be a multiple of the identity operator and the parity symmetry is not that trivial.
The fact that a pure state is a unit vector up to phases does not play any role here in my view. (It plays a crucial role in the proof of the Wigner theorem, so we have already used that fact.)
ADDENDUM. A deeper issue related with the definition of pure state in terms of vectors is the following one:
why are we allowed to choose the parity operator selfadjoint?
This fact is physically remarkable because it permits the parity to be interpreted as an observable.
The answer is like this. First of all, the parity operator $U$ is unitary or antiunitary as it is a symmetry. The choice depends on specific further requirements on that symmetry. For instance, dealing with non relativistic particles, the dual action of the parity symmetry $U$ on position and momentum is the natural one,
$$U X U^{-1}= -X\:, \quad U P U^{-1}= -P\:.$$
If $U$ were antiunitary the requirements above would be incompatible with the CCR
$$[X,P] = -i I\:.$$
The inclusion of the spin does not affect this result, though the analysis becomes more delicate.
From now on, we therefore assume that $U$ is unitary.
Evidently, and it is also generally stated in Wigner's theorem, $U$ can be re-defined with a multiplicative phase. We want to prove that we may fix the arbitrary phase of $U$ in order to make it selfadjoint.
For physical reasons, as $U$ represents the parity transformation, the action of $U^2$ on pure states has to be the identity. In other words, since states are unit vectors up to phases,
$$U^2\psi = c_\psi \psi\:.$$
Let us prove that, actually, $c_\psi$ does not depend on $\psi$.
Consider a Hilbert basis $\{\psi_j\}_{j\in J}$ and the further unit vectors $\phi_j= \frac{1}{\sqrt{2}}(\psi_{j_0} + \psi_j)$ where $j_0\in J$ is fixed and $J \ni j\neq j_0$.
By hypothesis
$$U^2 \phi_j = b_j \phi_j= b_j\frac{1}{\sqrt{2}}(\psi_{j_0} + \psi_j)\:.$$
On the other hand, since $U^2$ is linear,
$$U^2 \phi_j = \frac{1}{\sqrt{2}}(c_{j_0}\psi_{j_0} + c_j\psi_j)\:.$$
Comparing with the identity above:
$$(b_j- c_{j_0})\psi_{j_0}= (c_j-b_j)\psi_j\:.$$
Since $\psi_{j_0}$ and $\psi_{j}$ are orthogonal, we conclude that:
$$b_j = c_j= c_{j_0} =:c \quad \forall j \in J\:.$$
In summary $$U^2 \psi_j = c \psi_j \quad \forall j \in J\:.$$
Decomposing any vector along the basis $\{\psi_j\}_{j\in J}$, we conclude that
$$U^2 = cI$$
where $|c|=1$ as said above.
Defining
$$P:= c^{-1/2}U\:,$$
since $c^{-1/2}$ is still a unit complex number (defined up to a sign), we have that $P$ is still unitary, so that $P^*= P^{-1}$. But also that
$$PP=I\:.$$
The uniqueness property of the inverse operator eventually yields
$$P=P^*$$
and this is the wanted result: $P$ is simultaneously unitary and selfadjoint.
