I suggest you include some background. I would be surprised if more than one reader of MathOverflow knew (without doing some research) what the Ding projective dimension is (and I’m pretty sure that “Ding” should have a capital “D”). But I wouldn’t be surprised if more than one could answer your question if you gave them a bit more help.

@PeterKropholler In the paper of Faith that I linked to in a previous comment, he considers the following property (“FGTF”) of a ring $R$: every finitely generated submodule of a product of copies of $R$ embeds in a free module. This seems weaker than the property you want, but related. In the introduction, it is claimed that every left Noetherian ring is right FGTF, so maybe being Noetherian on the other side will be the relevant condition.

A relevant paper of Faith.

@PeterKropholler I think (or at least hope!) that I’ve edited my answer so that it is correct.

@PeterKropholler You make a good point! I think the example does work, but of course my justification is nonsense. I’ll delete the answer in a while if I don’t manage to fix it quickly.

The link is to Andrey Bovykin's (former) personal webspace at Bristol. If anybody has kept a copy, it's likely to be him. Have you tried contacting him? You can find his Gmail address by searching for him at UFBA.

For power maps, you don’t need $k\in\mathbb{N}$. Negative $k$ works as well.

Can’t you just take $f_i=x_ix_n$ and $G=x_1x_n$?

This works so long as $P$ is finitely generated, but if it is not, then $Q$ might not be contained in any maximal submodule of $P$. In fact, an infinitely-generated projective module may have a largest semisimple quotient (which can even be simple), but not be a projective cover of that quotient.