Note: $(d)$ is a nonzero proper ideal of $R$.
My attempt:
Consider $\pi:R\to R/(d)$, then $\ker(\pi)=(d)$. I want to prove that there is no $\sigma:R/(d)\to R$ such that $\pi\circ \sigma=id_M$.
Suppose that such a $\sigma$ exists, then $R\cong (d) \oplus R/(d)$. I'm not entirely sure if this is a contradiction. I know that the ideals of $R/(d)$ are the (principal) ideals of $R$ that contain $(d)$, but other than that I don't really see where I could use the fact that $R$ is a PID.
Thanks.
Starting from where you ended, $R\cong (d)\oplus R/(d) $ you could refine it to say that actually $R=(d)\oplus N$ (an internal direct sum). Since a domain has no proper summands, $(d)=\{0\}$ or $(d)=R$.
Both cases are just fine, since both $\{0\}$ and $R$ are projective, but presumably you wanted a nonzero, proper ideal $(d)$.
In fact, the same reasoning shows you $R/I$ is not projective for any nontrivial ideal $I$ in a domain, or even more generally, any uniform ring.
Another way:
A projective module over a PID is free, and a nonzero free module is faithful (i.e. has trivial annihilator) but $M$ is clearly annihilated by $d$. (Again, I exclude $(d)$ equal to $\{0\}$ or $R$, which I guess you omitted unintentionally.)