Borel sets modulo meagre sets and sets with the Baire property modulo meagre sets are isomorphic (because every set with the Baire property symmetrically differs from an open set by a meagre set, and every Borel set has the Baire property). The relationship between Borel sets and sets with the Baire property is the "Baire category" analogue of the relationship between Borel sets and Lebesgue measurable sets in measure theory.

Spectrum is another name for Stone space, and the name I more usually use (I must have lapsed into it without realizing that you might not know what it meant). So the spectrum of $B$ is a complicated non-metrizable space not homeomorphic to the Cantor space. By "we get $B$ back again", I mean that the map taking an $B$ to its corresponding equivalence class of Baire sets modulo meagre sets is an isomorphism (exactly the statement of Loomis-Sikorski).

In fact, for the spectrum of $B$, the two algebras differ - if we take the $\sigma$-algebra generated by the clopens (or equivalently the $\sigma$-algebra of Baire sets) modulo meagre sets, we get $B$ back again, and if we take Borel sets modulo meagre sets (or we could also use sets with the Baire property (a very different thing from Baire sets) modulo meagre sets), we get the completion of $B$.

Yes. You can find a proof of this in Halmos's Lectures on Boolean Algebras, section 13, theorem 4. But note that it is only because $2^\omega$ is metrizable that the $\sigma$-algebra generated by the clopen sets is the Borel $\sigma$-algebra.

Many operator algebraists don't like $E'$ for the dual of $E$ because this is used for the commutant, and one sometimes wants to refer to both the commutant of an algebra and its continuous dual.

Then $\phi^{-1}(1)$ is a non-principal ultrafilter on $X$. Therefore, if each set has a Stone-Čech compactification, every set has a non-principal ultrafilter. But in jstor.org/stable/2272118 Solovay and Pincus constructed a model of ZF + DC in which the Hahn-Banach theorem holds, but no set has a non-principal ultrafilter.

I know this is an old answer, but the second paragraph is not correct, over ZF (sans choice). If any infinite set $X$ has a Stone-Čech compactification (satisfying the universal property), as $X$ is not compact, there exists $y \in \beta X \setminus X$. For any $S \subseteq X$, $\chi_S : X \rightarrow 2$ can be lifted to $\tilde{\chi_S}$, as $2$ is a compact Hausdorff space. The map $\phi : \mathcal{P}(X) \rightarrow 2$ defined by $\phi(S) = \tilde{\chi_S}(y)$ can be proved to be a Boolean homomorphism vanishing on singletons using the uniqueness part of the universal property.

Ah yes, all diagrams in a poset commute. For an explict $X$ and $Y$ you could take, e.g. $\mathbb{I}, \mathbb{Q}$ as the irrationals and rationals respectively. Then $\mathbb{I} \cap \overline{\mathbb{Q}} = \mathbb{I} \cap \mathbb{R} = \mathbb{I}$, but $\overline{\mathbb{I} \cap \mathbb{Q}} = \overline{\emptyset} = \emptyset$.

IV. 6.1 in Schaefer's Topological Vector Spaces states that if $E$ is barrelled, $E'$ is quasicomplete for any $\mathfrak{S}$ topology defined by bounded sets, so in particular, the strong topology (as you define it, though not necessarily by Schaefer's definition of strong topology). So one direction is right.

The first statement is not correct without either having $S$ be complete as well, or $\mathcal{P}(S)$ being Radon probability measures instead of probability measures. A counterexample is a Lebesgue unmeasurable subset $S$ of $[0,1]$ with outer measure 1 equipped with the subspace topology - the restriction of Lebesgue measure to $S$ is a singleton in $\mathcal{P}(S)$, and therefore compact, but it is not tight.

@Alex Mennen: The natural measure-theoretic condition coming from the theory of the Giry monad ( link.springer.com/chapter/10.1007/BFb0092872 ) is that a bounded measurable function $P(X) \to \mathbb{R}$ arises from integration against a bounded measurable function $X \to \mathbb{R}$ iff $$ f(\mathrm{proj}(\mathbb{P})) = \int_{P(X)}f d\mathbb{P} $$ for all $\mathbb{P} \in P(X)$. Is the condition you give in the question intended to rephrase this condition, or did you come up with it independently?

It is the boundedness of the set of probability measures that I am using, not the positivity (so $\langle -, f\rangle$ is also continuous on the set of signed measures of variation $\leq 1$). I do not know if $\langle -, f \rangle$ will be continuous when restricted to positive measures.

I realize I forgot to address the part of the question about the $\sigma$-algebra on $P(X)$. You can get some of the distance with that by using the theory of the Giry monad.