You don't need projectivity to define the module structure. However if you want your duality to satisfy something like $\mathrm{Hom}(\mathrm{Hom}(B,R),R)\cong B$ then $B$ being a finitely generated projective left $R$-module is sufficient for $\mathrm{Hom}$ as I defined it.

You mean the left $R$-module action I think. $((rs)\cdot f)(b)=f(brs)=(s\cdot f)(br)= (r\cdot(s\cdot f))(b)$

There isn't enough context in the question but if you know that $B$ is a projective left $R$-module and nothing else there is a fair chance that you want $\mathrm{Hom}(B,R)$, where $\mathrm{Hom}$ means left $R$-linear maps, with left $R$-module structure coming from the right $R$-module structure on $B$ --- $(r\cdot f)(b)=f(br)$ --- and right $R$-module structure coming from the right $R$-module structure on $R$ --- $(f\cdot r)(b) = f(b)r$.

How do you define quasi-coherent sheaves in this context?

I don't think that the spectral stabilises until the third page: there can be (indeed given YCor's answer there are) non-zero maps $d\colon E^2_{i,0}\to E^2_{i-2,1}$.

Of course this also addresses Victor Protsak's question on another answer.

Agreed. In fact I think that when $P=I$, the augmentation ideal which is the important case here, the generators of $L(P)$ are precisely the weights of adjoint representation. I suppose it remains possible that the bad $\lambda$ are all a linear combination of at most $\dim \mathfrak{g}$ of these weights with repeats allowed (although the converse is not true). Perhaps even $H^i(\mathfrak{g},\mathbb{C}_\lambda)$ is zero unless $\lambda$ is a linear combination of $i$ of the weights (but not conversely).

If I were in the mood for making wild conjectures I might speculate that all homology groups of $\mathbb{C}_\lambda$ are trivial precisely if $\lambda$ is a sum of at most $\dim \mathfrak{g}/[\mathfrak{g,g}]$ of the generators of $L(I)$ given by Brown (if the sign was wrong before it is also wrong here). However I have no evidence for this.

It is possible that I've made sign errors near the end and that I mean that $(\mathbb{C}_{-\lambda})_S=0$ unless $\lambda\in L(P)\cup \{0\}$ and then $L(P)$ is the negation of what I claimed in the specific example.

If you want more details I can supply them another time. You can find the general idea in section 3 in one of my papers dpmms.cam.ac.uk/~sjw47/Euler.pdf but for another similar context.

Using this paper numdam.org/article/CM_1984__53_3_347_0.pdf of Brown I think you can easily get sufficient conditions for all homology groups to vanish. The idea I have in mind is that you can localise $U(\mathfrak{g})$ at the maximal Ore set $S$ contained in the complement of the augmentation ideal (described in the paper) and then compute homology by taking $(\mathbb{C}_S\otimes^L_{U(\mathfrak{g})_S} (\mathbb{C}_{\lambda})_S)$. $(\mathbb{C}_{\lambda})_S$ will already be zero except for on an explicitly parameterised set of $\lambda$.

There is a natural ring homomorphism $\theta\colon \mathbb{Z}_p[[G]]\to \mathbb{Z}_p$. Since ring homomorphisms send units to units it might help to consider the image of your elements under $\theta$.

I think that, as often, the motivation comes from the examples. i.e. all three conditions hold in a large family of interesting examples of $U(\mathfrak{g})$-modules.

For the definition of locally finite you should replace 'finite (as sets)' by 'finite dimensional (as vector spaces)' in your boldface definition. Equivalently every element of $M$ lives in a $U(\mathfrak{p})$-submodule that is finite dimensional as a vector space.