Unfortunately, you will run into a basic difficulty in your desire to make this example concrete, which is that the canonical bundle on projective space is anti-ample, and so the canonical bundle is not effective. (Rather, it is its dual that is an effective divisor.)
More precisely, if $H$ is the linear equivalence class of a hyperplane in $\mathbb P^n$, then the canonical divisor $K$ is equal to $-(n+1)H$.
So if you want to intersect $K$ with $V(P)$ (the variety cut out by the polynomial $P$) you will have to use at least a little bit of theory, even if only to interpret what you are doing.
You might be better of starting with $H$ itself. Then $H \cap V(P)$ is the linear equivalence class of a hyperplane section of $V(P)$. Assuming that $V(P)$ is smooth, then Bertini's theorem says that a generic member of the linear equivalence class $H\cap V(P)$ will be smooth, and even irreducible if the dimension of $V(P)$ is at least two (i.e. if $n \geq 3$).
Then one way to write $K \cap V(P)$ is simply as $-(n+1) \bigl( H \cap V(P) \bigr)$.
Alternatively, one could consider the linear equivalence class $(n+1)\bigl( H \cap V(P)\bigr).$ This is the class of intersections of $V(P)$ with a degree $n+1$ hypersurface, and again a generic member is smooth (and irreducible if $V(P)$ is of dimension at least $2$). Then you think of $K \cap V(P)$ as being
the negative of this class.