Let $k$ any field and $Y \subset \mathbb{P}^n_k$ closed subscheme. We are going to study the Hilbert polynomials in reversal sense:
Let recall a Hilbert polynomial associated to a coherent sheaf $F$ on $X=\mathbb{P}^n_k$ together with line bundle $O_X(1)$ is a polynomial
$$\Phi(t) = \sum_{i=0} ^n (-1)^i h^i(X, F \otimes O_X(1)^{\otimes t} )= \sum_{i=0} ^n (-1)^i h^i(X, F (t)) \in \mathbb{Q}[t] $$
The Hilbert polynomial associated to a closed subscheme $Y \subset \mathbb{P}^n_k$ as in our case defined by homogeneous ideal sheaf $I_Y$ it the Hilbert polynomial of the quotient sheaf $O_Y= O_X/I_Y$.
Two examples $Y \subset \mathbb{P}^n_k$ can be easily computed:
(i) $Y \cong \mathbb{P}^r_k$ embedded linearly in $\mathbb{P}^n_k$ over $k$ has Hilber polynomial $\Phi^Y(t)= \binom{r+ t}{r}$
(ii) $Y$ is a hypersurface $H_d$ of degree $d$ in $\mathbb{P}^n_k$ defined by a homogeneous polynomial $f \in k[x_0,...,x_n]$ obtains HP $\Phi_d^Y(t)= \binom{n+ t}{n}- \binom{n-d+ t}{n}$
The (ii) follows easily from exact sequence $ 0 \to O_{\mathbb{P}^n_k}(-d) \to O_{\mathbb{P}^n_k} \to O_{H_d} \to 0 $, the additivity of Hilbert polynomials for exact sheaf sequences and previous case (i).
Now the question is how to show that cases (i) & (ii) can be reversed:
(i) If closed subscheme $Y \subset \mathbb{P}^n_k$ has Hilbert polynomial $\Phi^Y(t)= \binom{r+ t}{r}$, why $Y \cong \mathbb{P}^r_k$ and embedded linearly in $\mathbb{P}^n_k$ over $k$?
(ii) if $Y \subset \mathbb{P}^n_k$ has Hilbert polynomial $\Phi_d^Y(t)= \binom{n+ t}{n}- \binom{n-d+ t}{n}$, why $Y$ is a hypersurface $H_d$ of degree $d$.
On (ii) the paper by Nitsure on the construction of Hilbert scheme that originally motivates this question has a hint (page 6): If $Y \subset \mathbb{P}^n_k $ is a closed subscheme with Hilbert polynomial of degree $n − 1$, then show that the schematic closure $Z$ of the height $1$ primary components is a hypersurface in $\mathbb{P}^n_k$ with $\operatorname{deg}(Z) = \operatorname{deg}(Y)$.
Contextually the "height $1$ primary components" of $Y$ seems to be the irreducible components of $Y$ of codimension $1$. Why the author called it "primary" I not know. The Zariski topology don't draw a difference between prime ideals and primary ideals having primes as their radicals. Nevertheless I assume that the hint refers to irreducible components $C_{i, max} \subset Y$ of maximal dimension.
Set $Z:= \overline{\bigcup_i C_{i, max}}$ and the claim is $\operatorname{deg}(Z) = \operatorname{deg}(Y)$. Why does this equality hold? And why this implies that $Y$ is hypersurface of degree $d$?
Here are two important facts about Hilbert polynomials. If $Y\subset\mathbb{P}^n$, and its Hilbert polynomial is $a_0t^r+a_1t^{r-1}+\cdots+ a_r$, with $a_0\neq 0$ (so degree is $r$), then $\dim Y=r$ and degree of $Y=r!a_0$.
Both are easy to prove. If $r=0$, I will leave you to check that this is true. So, assume $r>0$. Then it is easy to see that $\dim Y>0$ and we have an exact sequence, taking a general hyperplane section, $$0\to O_Y(-1)\to O_Y\to O_{Y\cap H}\to 0.$$ Now, by additivity, one can calculate the Hilbert polynomial of $Y\cap H$ and use induction to finish the proof.
This should answer (i) and (ii).