Let $M$ lie in $\operatorname{mod}\Lambda$ for a finite dimensional $k$-algebra $\Lambda$.
Let $P = (P_{1} \xrightarrow{d} P_{0})$ be a silting complex in $K^{b}(\operatorname{proj}\Lambda)$ and $M = \operatorname{Cok}(d)$.
So $P_1 \xrightarrow{d} P_{0} \xrightarrow{\pi} M$ is a projective presentation of $M$. Then, one has a decomposition $P_1 = P_{1}' \oplus P_{1}''$ such that $d' := d|_{P_{1}'}$ is right minimal and $d|_{P_{1}''} = 0$.
In $\tau$-tilting by T. Adachi, O. Iyama and I. Reiten, in the proof of proposition 3.6(b), they say that then $$P_{1}' \xrightarrow{d'} P_{0} \xrightarrow{\pi} M$$ is a minimal projective presentation. I can't see why $\pi$ needs to be right minimal, and thus a projective cover. Any help?
Attempt (following Jeremy's answer):
So, let $d \in \operatorname{rad}(P_{1}, P_{0})$ and suppose that $\pi$ is not right minimal. We then have a decomposition
$$\pi = (\pi', 0) : P_{0}' \oplus P_{0}'' \rightarrow M$$
such that $\pi'$ is right minimal. By composing $d$ with the projections of $P_{0}$ onto its two summands $P_{0}', P_{0}''$, calling them $d', d''$ we have the sequence
$$P_{1} \xrightarrow{\begin{pmatrix} d' \\ d''\end{pmatrix}} P_{0}' \oplus P_{0}'' \xrightarrow{(\pi',0)} M \rightarrow 0$$
Further supposing that $d = \begin{pmatrix} d' \\ d''\end{pmatrix}$ is not right minimal, we have a decomposition
$$P_{1}' \oplus P_{1}'' \xrightarrow{\begin{pmatrix} \tilde{d'} & 0 \\ \tilde{d''} & 0\end{pmatrix}} P_{0}'\oplus P_{0}'' \xrightarrow{(\pi', 0)} M$$
such that $\begin{pmatrix} \tilde{d'} \\ \tilde{d''}\end{pmatrix}$ is right minimal. Then $\pi d = \begin{pmatrix}\pi' & 0 \end{pmatrix}\begin{pmatrix} \tilde{d'} & 0 \\ \tilde{d''} & 0\end{pmatrix} = \pi' \tilde{d'}$.
Does this imply that $\operatorname{Im}(d) = \operatorname{Im}(\tilde{d'})$? If so, then I think I'm able to show that $P_{1}'$ is isomorphic to $P_{0}''$ by the restriction of $d$, thus contradicting the property of $d$ being in the radical, as explained by Jeremy.
As remarked just after Definition 3.1 of that paper, any two-term complex of projectives is isomorphic (in $K^b(\text{proj }\Lambda)$) to a "minimal" one, where the differential is in the radical. I suspect that in the passage you refer to, the authors are implicitly assuming that they have chosen a minimal representative of the isomorphism class. If not, then it's not true that $\pi$ must be right minimal.