Let $P_0=|\psi\rangle\!\langle\psi|$ and $P_1=I-P_0$. This is a projective measurement which deterministically distinguishes the two orthogonal states.
More generally, consider a projective measurement with operators $\newcommand{\ket}[1]{\lvert#1\rangle}\{P_i\}_{i=1}^d$ and $\newcommand{\braket}[2]{\langle #1\rvert #2\rangle}\newcommand{\ketbra}[1]{\lvert #1\rangle\!\langle #1\rvert} P_i\equiv\ketbra{\eta_i}$ where $\braket{\eta_i}{\eta_j}=\delta_{ij}$, and a set of (not necessarily orthogonal) states $\{\ket{\psi_i}\}_{i=1}^\ell$ with $\ell\le d$.
To distinguish the states deterministically we need
$$\operatorname{Tr}(P_i \ketbra{\psi_j})=|\braket{\eta_i}{\psi_j}|^2=\delta_{ij}.\tag2$$
Define the matrices $\Pi\equiv\sum_{i=1}^d |\eta_i\rangle\!\langle i|$ and $\Psi\equiv\sum_{i=1}^\ell|\psi_i\rangle\!\langle i|$. Note that $\Pi$ is $d\times d$ and $\Psi$ is $d\times\ell$.
Eq. (1) is thus equivalent to $\Pi^\dagger \Psi=I_{d\times \ell}$ (we can tune the definitions of the states $\ket{\eta_i}$ to have $\braket{\eta_i}{\psi_i}\in\mathbb R$ without any loss in generality).
This is only possible if $\Psi$ is "maximally entangled", i.e. has rank $\ell$ and all its (nonzero) singular values equal $1$ (equivalently, iff $\Psi^\dagger\Psi=I_{\ell}$). This is true iff the states $\ket{\psi_i}$ are orthonormal.
The last statement follows from observing that $\Psi^\dagger\Psi$ has the same nonzero singular values/eigenvalues as $\Psi\Psi^\dagger=\sum_{i=1}^\ell \ketbra{\psi_i}$. The latter having an $\ell$-fold degenerate eigenvalue $+1$ means
$$\Psi\Psi^\dagger = \sum_i\ketbra{\psi_i}=\sum_i\ketbra{\phi_i}$$
for some orthonormal set $\{\ket{\phi_i}\}_{i=1}^\ell$. This in turns implies that for some unitary $U$ we have $\ket{\psi_i}=\sum_j U_{ij}\ket{\phi_j}$, and thus $\braket{\psi_i}{\psi_j}=\delta_{ij}$.
This shows that non-orthogonal states cannot be distinguished deterministically via projective measurements (in fact, they cannot be distinguished deterministically by any measurement, but that's not what we are showing here).
