First of all, nowadays, one learns it logic courses that "truth" is a complex notion, and in particular that "truth" per se has no real meaning; only "truth within a certain domain of interpretation" has one. Therefore, a sentence like "$b$ is true" has to be interpreted in some way, and I hope it is in the way that I stated my answer below.
Moreover, in logic, when dealing with something to be thought as a "set of propositions" (such as a lattice, a Boolean algebra, etc.), it is kind of a subtelty to distinguish between formal equality of two formulas and some kind of equivalence between them. For example, when one tries to define logic in full rigor, one first defines strings of characters, and at this low level, for example, one can consider that $(a\wedge b) \neq a\wedge b$. The highest (on the scale of abstraction) level of equivalence is to say that two propositions $a$ and $b$ are equivalent if $a\Leftrightarrow b$ is provable, within some implicit logical framework. It turns out that for quite a number of logics, this equivalence is the same as semantic equivalence, for which two propositions $a$ and $b$ are equivalent if, within every possible domain of interpretation of the formulas, they are both true or both false. Since <PERSON> talks about "truth" and "falseness", I guess that it is correct to interpret his claim as stating a sort of semantic equivalence.
Let $L$ be a set, $\wedge,\vee : L\times L \rightarrow L$. Call a map $t: L \rightarrow {\top,\bot}$ a truth assignment if $\forall a,b \in L,\ t(a\wedge b) = \top \Leftrightarrow t(a) = t(b) = \top$ and $\forall a,b \in L,\ t(a\vee b) = \top \Leftrightarrow \left(t(a) = \top \ \textbf{or}\ t(b) = \top\right)$.
We say that two elements $a,b \in L$ are semantically equivalent if for all truth assignment $t$, $t(a) = t(b)$. Note that in classical logic, semantic equivalence is the same as logical equivalence.
Proposition: For all $a,b,c \in L$, $a \wedge (b \vee c)$ and $(a\wedge b) \vee (a\wedge c)$ are semantically equivalent.
Proof: Just examine all the cases! I'll look at one case, as an example. 330
Let $t$ be a truth assignment. Assume that $t(a \wedge (b \vee c)) = \top$. Then $t(a) = \top$ and either $t(b) = \top$ or $t(c) = \top$. Assume $t(b) = \top$. Then $t(a\wedge b) = \top$, then $t((a\wedge b)\vee (a\wedge c)) = \top$. We can argue similarly when $t(c) = \top$.
As for the diagram, it represents the lattice of propositions on a two-dimensional Hilbert space. One can assume that this space represents the system consisting in a single photon, and every nontrivial proposition is of the form "the photon goes out of the polarizer with some angle". If one denotes $a_\phi$ the proposition "the photon goes out of the polarizer with angle $\phi$", what does $a_{\phi\ or\ \phi'}$ mean? Should it mean "the photon goes out of the polarizer with angle $\phi$ when put in, or goes out of the polarizer with angle $\phi'$ when put in"? Such a proposition does not exist: for opposite angles $\phi$ and $\phi'$, such a proposition would be greater or equal than $a_\phi \vee a_{\phi'} = true$, so would be $true$. However, any photon that goes out of a polarizer with a different angle would sometimes, in one of the $\phi$ and $\phi'$ polarizers, not go out, so $a_{\phi \ or \ \phi'}$ cannot be a valid quantum proposition. 330