How well defined is $log(P)$ for $P$ projection?

$P$ is a projection operator in the limiting case where $P$ represents a state that is completely known, i.e. a pure state, so the entropy is zero. As a limiting case, the valid and well-defined mathematical description is $$lim_{P rightarrow 0^+} P log(P)=0.$$ This is still a bit sloppy. Since $P$ is a matrix, we're actually taking the trace and $P rightarrow 0^+$ means as P goes to a representation of a pure state. It's a bit more clear to consider entropy in terms of the eigenvalues of $P$, $$S=sum lambda_i log lambda_i,$$ so that there is no ambiguity in interpreting $$lim_{lambda rightarrow 0^+} lambda log(lambda)=0,$$ for any given term in the sum.

In response to your comment, when $P$ is a projection operator, the expression $log(P)$ is inherently undefined. You can see this by noting that projection operators are idempotent by definition, so $P^2=P$.

Consider the proposition that we can express $P$ in exponential form $P=e^x$ for some unknown $x$. The idempotentcy of $P$ requires $e^x=e^{2x}$, which tells us that $x$ is imaginary with magnitude $2 pi n, , n in mathbb{Z}$, over some $d$-dimensional basis.

Whatever appropriately normalized basis we pick, say $b_d$, the inevitable result is $e^{i2pi n b_d}=I_d$. So we get back the identity instead of our projection operator, contradicting the proposition.

The is one of several ways to see that we can't define $P$ as an exponential, and that $log(P)$ is hopelessly undefined.