When considering the multplication endomorphism \begin{equation*} \begin{split} [\times z]_{K/Q}: & \:\: K \rightarrow K \\ & \:\: x \:\mapsto xz \end{split} \end{equation*} for $Q$ a field and $K$ a finite $Q$-algebra, we get the eigenvalues $z_1,...z_n$ which are the roots of $P_{K/Q,car,z}(X) = \text{det}(X.\text{Id}_K-[\times z])$.
Now the course notes that I'm working with quickly mention that the eigenvalues of $P([\times z]_{K/Q})$ are $P(z_1),...,P(z_n)$ for any polynomial $P\in Q[X]$ and I'm really not convinced.
A first thing to understand here is that the result of substituting $[\times z]_{K/Q}$ for $X$ into a polynomial $P$, which result your write as $P([\times z]_{K/Q})$ (I would personally prefer the more explicit $P[X:=[\times z]_{K/Q}]$) is equal to to operation of multiplication by $P(z)$ (more explicitly by $P[X:=z]$), so in formula $$ P([\times z]_{K/Q})=[\times P(z)]_{K/Q}. $$ To see this, there is really not much more than to observe that $([\times z]_{K/Q})^k(y)=z^ky$ for any $y\in K$ and any exponent$~k$, and then use linearity to pass from these monomials to general polynomials.
Now over a suitable extension of $Q$, one can find a basis such that $[\times z]_{K/Q}$ is represented on it by an upper triangular matrix with diagonal entries $z_1,\ldots,z_n$ (if they are all distinct one can even get a diagonal matrix, which makes intuition a bit simpler). On that basis $P([\times z]_{K/Q})$ is upper triangular matrix with diagonal entries $P(z_1),\ldots,P(z_n)$, and the result follows.
(I realise that I did not even use what is said in the first paragraph, but I'll let it since it seems relevant anyway.)