Image of Kähler class of an octic K3 surface is primitive and satisfies $p(\kappa) \cdot p(\kappa) = 8$

Question

I am trying to understand the following sentence from a paper by Kovalev called Constructions of Compact $G_2$ Holonomy Manifolds. Unfortunately I can't find a copy of the paper publicly available so I will try to give the context. Here is the sentence:

"If $D \subset \mathbb{C}P^5 $ is an octic K3 surface then the image $p(\kappa_I)$ of the Kähler class of $D$ is primitive and $p(\kappa_I) \cdot p(\kappa_I)=8$"

1. $D$ being "octic" means it is degree 8. In this case, $D$ is the complete intersection of three quadrics in $\mathbb{C}P^5$.
2. The map $p:H^2(D,\mathbb{Z}) \rightarrow L$ is an isomorphism of lattices where $L$ is the so-called K3 lattice and $L = 3H \oplus 2(-E_8)$. Here, the $H$ is the lattice with bilinear form $\begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}$. This is a lattice that has signature $(3,19)$ and such an isomorphism (called a marking) always exists for K3 surfaces.
3. The Kähler class $\kappa_I$ has been normalized "according to the volume form", which I assume means that $\kappa_I \wedge \kappa_I =$ volume form for $D$ and $\int_D \kappa \wedge \kappa = 1$.
4. The dot in $p(\kappa_I) \cdot p(\kappa_I)=8$ is the bilinear form for the lattice.
5. I am not sure if it matters, but I would like to take the Kähler metric associated to $\kappa_I$ to be the Ricci-flat one.
6. Apparently $p(\kappa_I)$ is also primitive, i.e. not divisible by an integer $\neq 1$ in $L$. This is another fact I am not sure how to prove.

My question: I have absolutely no idea where the number 8 comes from. I can see, from the properties of the lattice, that we must have an even number, but I am lost why it has to be 8, or is there a choice that was made somewhere that I don't know about? I will be happy to give more details from the paper if it is necessary.

EDIT: After some reading, I have found some information that might point to an answer but I can't quite put it all together.

In Griffiths and Harris, on page 191, there is a statement of the Kodaira Embedding Theorem: A compact, complex manifold $M$ is an algebraic variety -i.e., is embeddable in projective space- if and only if it has a closed, positive, $(1,1)$ - form $\omega$ whose cohomology class $[\omega]$ is rational.

Now, if $D$ is an octic K3 surface, then it certainly has a positive $(1,1)$-form since in particular, it is Kähler and I believe I can guarantee that this form is integral, not just rational by scaling the volume form on $D$. Thus, by the proof of the theorem on page 191, there exists a holomorphic line bundle $L \rightarrow M$ with $c_1(L) = [\omega]$. I am guessing that this means that the Kähler class does indeed "correspond in some sense" to a very ample divisor that embeds $D$ in $\mathbb{C}P^5$ as Tabes Bridges said in the comment below. Unfortunately my algebraic geometry is too weak to really understand.