Is it true that the number is divisible by $p$?

The answer is yes, and I will assume that you are familiar with modular arithmetic. First write $$P(k)=a((2ak+b)^2-(b^2-4ac)),$$ and note that $2ak+b$ ranges over all residue classes modulo $p$ as $k$ ranges from $x+1$ to $x+2p-1$, because $p$ doesn't divide $a$. It follows that $(2ak+b)^2$ ranges over all quadratic residues modulo $p$.

Given that $P(k)$ is a square for all $k$ ranging from $x+1$ to $x+2p-1$, and that the map on residue classes mod $p$ $$(\Bbb{Z}/p\Bbb{Z})\ \longrightarrow\ (\Bbb{Z}/p\Bbb{Z}): n\ \longmapsto a(n-\Delta),$$ is bijective, again because $p$ does not divide $a$, we see that this map restricts to a bijection on the quadratic residues modulo $p$. In particular $P(k)\equiv0\pmod{p}$ for some $k\in\{x+1,\ldots,x+2p-1\}$, from which point your argument works.

From this point, one could also note that there are more quadratic residues than quadratic nonresidues. Depending on whether $a$ is a quadratic residue or not, the map $$(\Bbb{Z}/p\Bbb{Z})\ \longrightarrow\ (\Bbb{Z}/p\Bbb{Z}): n\ \longmapsto n-\Delta,$$ either maps all quadratic residues to quadratic residues, or maps all quadratic residues to quadratic nonresidues. As this is a bijection, the latter is impossible and so $a$ is a quadratic residue. Moreover, if $\Delta\not\equiv0\pmod{p}$ then repeated application of the map above implies that all residue classes modulo $p$ are quadratic residues, contradicting the fact that $p>3$. Hence $\Delta\equiv0\pmod{p}$, i.e. $b^2-4ac$ is divisible by $p$.

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Note that for this argument to work, it suffices for only $p$ consecutive integers to exist such that $P(x+1)$ through $P(x+p)$ are squares. Or even more weakly, but less elegantly, for $P$ to take on a square value at least once in each residue class modulo $p$.