Let $\boldsymbol\mu\equiv\{\mu_j\}_{j=1}^m\subset \operatorname{Pos}(\mathbb{C}^d)$ be a POVM whose elements act on a $d$-dimensional space and have unit rank, that is, $\mu_j = w_j \mathbb{P}_{\psi_j}$, $\mathbb{P}_\psi\equiv|\psi\rangle\!\langle\psi|$, for some set of pure states $|\psi_j\rangle$, and non-negative coefficients $w_j$ such that $\sum_j w_j=d$.
Note that we must have $m\ge d$, as otherwise there's not enough elements to obtain a rank-1 POVM.
Then you can implement this POVM as a projective measurement on an $m$-dimensional space, by taking the isometry $V$ such that $V^\dagger |j\rangle=\sqrt{w_j}|\psi_j\rangle$, that is,
$$V = \sum_j \sqrt{w_j} |j\rangle\!\langle \psi_j\rvert.$$
Notice that with this you have by definition $\mu_j = V^\dagger \mathbb{P}_j V = \mathbb{P}(V^\dagger|j\rangle)$. In words, we're saying that the vectors the POVM elements project onto are (modulo scalar factors and a complex conjugate) the rows of $V$.
For example, consider the single-qubit SIC-POVM, whose elements are
$$\mu_1 = \frac12\mathbb{P}_0,
\qquad \mu_2 = \frac12 \mathbb{P}\left(\frac{|0\rangle+\sqrt2|1\rangle}{\sqrt3}\right), \\
\mu_3 = \frac12 \mathbb{P}\left(\frac{|0\rangle+\sqrt2 \omega_3 |1\rangle}{\sqrt3}\right),
\qquad
\mu_4 = \frac12 \mathbb{P}\left(\frac{|0\rangle+\sqrt2 \omega_3^2 |1\rangle}{\sqrt3}\right).
$$
You can implement this via an isometry $V:\mathbb{C}^2\to\mathbb{C}^4$, which you can imagine sending the input qubit to two output qubits, and has the form
$$V = \frac1{\sqrt6}\begin{pmatrix}\sqrt3&0 \\ 1 & \sqrt2 \\ 1 &\sqrt2 \omega_3 \\ 1 & \sqrt2 \omega_3^2\end{pmatrix}.$$
Notice how this allows to implement the POVM using only two qubits, in contrast with Naimark's theorem, which would give you a three-qubit strategy instead, as Narimark's isometry would in this case have the form
$$V_{\rm Naim} = \frac{1}{3\sqrt2}\begin{pmatrix}3&0\\0&0 \\ 1 & \sqrt2 \\ \sqrt2 & 2 \\ 1&\sqrt2 \omega_3^2 \\ \sqrt2\omega_3 & 2
\\ 1&\sqrt2 \omega_3 \\ \sqrt2\omega_3^2 & 2.
\end{pmatrix}$$
