Suppose you are given either $\rho_1$ or $\rho_2$, and you also know that the probabilities you got one or the other are $p_1$ and $p_2$, respectively. If you have no prior knowledge of the frequencies with which I'm going to give you one state or the other, you just use $p_1=p_2=1/2$.
You are asking what's the measurement that optimally distinguishes between $\rho_1$ and $\rho_2$, given the priors $p_1,p_2$.
Consider a simple scheme where the measurement outcome is directly used to decide which one was the input state.
We are thus looking at two-outcome measurements, i.e. POVMs with elements $\boldsymbol\mu\equiv \{\mu_1,\mu_2\}$, and use the strategy of guessing $\rho_i$ if the $i$-th outcome is found.
Note that there is no loss of generality in fixing this strategy: you could consider a more general scenario where upon finding the outcome $i$, some function $f:\{0,1\}\to\{0,1\}$ is used to determine the input state, but any such strategy just amounts to a relabelling of the POVM elements.
The (average) success probability of this strategy, when using the measurement $\boldsymbol\mu$, reads
$$p_{\rm success}(\boldsymbol\mu) \equiv \sum_i \langle\mu_i, p_i\rho_i\rangle
= \langle\mu_1, p_1\rho_1\rangle
+ \langle\mu_2, p_2\rho_2\rangle.$$
Using the normalisation condition on the POVM, $\mu_1+\mu_2=I$, we get
$$p_{\rm success}(\boldsymbol\mu) = p_2 +
\langle\mu_1, p_1\rho_1- p_2\rho_2\rangle.
$$
We get the overall optimal success probability maximising over all such POVMs:
$$\mu_{\rm optimal} \equiv \operatorname*{argmax}_{\boldsymbol\mu} p_{\rm success}(\boldsymbol\mu)
= \operatorname*{argmax}_{\boldsymbol\mu}
\langle\mu_1, p_1\rho_1- p_2\rho_2\rangle.$$
To get the optimal success probability, observe that
$$\max_{\boldsymbol\mu} p_{\rm succ}(\boldsymbol\mu)
= p_2 + \max_{0\le \mu_1\le I} \langle \mu_1,p_1 \rho_1-p_2\rho_2\rangle.$$
To compute this maximum, observe that for any Hermitian $P$,
$$\max_{0\le \mu_1\le I} \langle \mu_1,P\rangle = \operatorname{tr}(P_+) = \frac12(\|P\|_1 + \operatorname{tr}(P)),$$
where $P_\pm$ are the unique positive definite operators such that $P_+ P_-=0$ in the decomposition $P=P_+ - P_-$, and $\|P\|_1=\operatorname{tr}(P_+)+\operatorname{tr}(P_-)$. The maximum is achieved with $\mu_1$ the projection onto the support of $P$: $\mu_1=\Pi_{\operatorname{supp}(P)}$.
We thus have
$$\max_{0\le \mu_1\le I} \langle \mu_1,p_1 \rho_1-p_2\rho_2\rangle
= \frac12((p_1-p_2) + \|p_1 \rho_1-p_2\rho_2\|_1),$$
and
$$\max_{\boldsymbol\mu} p_{\rm succ}(\boldsymbol\mu)
= \frac12 + \frac12\|p_1 \rho_1-p_2\rho_2\|_1,$$
with the optimal measurement being the POVM with $\mu_1$ projection onto the support of $p_1\rho_1-p_2\rho_2$, and $\mu_2=I-\mu_1$ projection onto its orthogonal complement.
More explicitly, this means that the optimal measurement can always be chosen to be projective, and corresponding to measuring in the eigenbasis of $p_1\rho_1-p_2\rho_2$.
It is still possible to have non-projective optimal measurements, for example by adding projections onto the ker of $p_1\rho_1-p_2\rho_2$ to the POVM elements. Any such change won't affect the state discrimination procedure in any way, but it can give non-projective optimal POVMs (if arguably in a somewhat trivial manner).
Much more information can be found in https://arxiv.org/abs/1707.02571.
As an example, let $p_1=p_2=1/2$ and
$$\rho_1=\frac23\begin{pmatrix}1/2 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0&0\end{pmatrix},
\qquad
\rho_2=\frac23\begin{pmatrix}1/2 & 0 & 0 \\ 0 & 1/2 & 1/2 \\ 0 & 1/2 & 1/2\end{pmatrix}.$$
Then, the nonzero eigenvalues of $\Delta\rho$ are $\lambda_\pm=\pm\sqrt2/3$, and a possible non-projective optimally distinguishing two-outcome POVM is
$$\mu_1 = |\lambda_+\rangle\!\langle\lambda_+| + \frac12 |0\rangle\!\langle0|, \\
\mu_2 = |\lambda_-\rangle\!\langle\lambda_-| + \frac12 |0\rangle\!\langle0|.$$
