$\newcommand\la\lambda$
Let $g(z)$ denote some branch of $\sqrt z$, so that $Q(z)=P(z)^2-g(z)^2$ and $P^{(k)}(\la_j)=g^{(k)}(\la_j)$ for each $j$ and all $k=0,\dots,m_j-1$, where the $\la_j$'s are the distinct eigenvalues of $A$ and the $m_j$'s are their multiplicities. Then, by the Leibniz rule, $Q^{(k)}(\la_j)=0$ for each $j$ and all $k=0,\dots,m_j-1$.
So, for some polynomial $R(z)$, \begin{equation} Q(z)=R(z)\prod_j(z-\la_j)^{m_j} =R(z)C(z), \tag{1}\label{1} \end{equation} where $C(z):=\prod_j(z-\la_j)^{m_j}$ is the characteristic polynomial of $A$ (see details on the first equality in \eqref{1} below). So, the polynomial $Q(z)$ is indeed a multiple of the characteristic polynomial of $A$.
(The previous answer, referred to in your question, does not even mention indices.)
Details on the first equality in \eqref{1}:
Lemma 1: Let $q(z)$ be a polynomial such that $q^{(k)}(\la)=0$ for some $\la$ and all $k=0,\dots,m-1$, where $m$ is an integer $\ge1$. Then $(z-\la)^m$ is a divisor of the polynomial $q(z)$ -- that is, $q(z)=(z-\la)^m S(z)$ for some polynomial $S(z)$.
Proof of Lemma 1: By shifting, without loss of generality $\la=0$. Divide $q(z)$ by $z^m$ with a remainder $r(z)$ of degree $\le m-1$, so that $q(z)=z^m s(z)+r(z)$ for some polynomial $s(z)$. Then for all $k=0,\dots,m-1$ we have $0=q^{(k)}(0)=r^{(k)}(0)$. Because $r(z)$ is of degree $\le m-1$, it follows that $r(z)$ is the zero polynomial, so that $q(z)=z^m s(z)$, which completes the proof of Lemma 1. $\quad\Box$
Now, by part 1 of the answer, $Q^{(k)}(\la_j)=0$ for each $j$ and all $k=0,\dots,m_j-1$. So, for each $j$, by Lemma 1, the polynomial $(z-\la_j)^{m_j}$ is a divisor of the polynomial $Q(z)$. Also, the polynomials $(z-\la_j)^{m_j}$ are coprime for different values of $j$, since the $\la_j$'s are distinct. So, $\prod_j(z-\la_j)^{m_j}$ is a divisor of $Q(z)$; that is, the first equality in \eqref{1} holds.
It remains to provide