Romanian Mathematical Olympiad

As $k$ and $p$ are odd, $|K|$ is even, in fact a power of $2$, the characteristic of $K$ being $2$. It follows that $x_1, x_2, \dots, x_t$ are the roots of $P$. Consider $P = \sum_{i=0}^{t} a_j X^j$.

The multiplicative group $(K^*, \cdot)$ is cyclic of order $kp$. Consider $a \in K^*$ one of its generators. For any $s \in \{1, 2, \dots, kp\}$, with $(s, kp) = 1$ we get
$$\mathrm{ord}(a^s) = \mathrm{ord}(a) = kp. \qquad (1)$$
In particular, for any $i \in \{1, 2, \dots, p-1\}$ we have $(ik + p, k) = (p, k) = 1$ and $(ik + p, p) = (ik, p) = 1$, so that $(ik + p, kp) = 1$ and $\mathrm{ord}(a^{ik+p}) = kp$. These give

$$A_r = \sum_{i=0}^{p-1} a^{-irk} P(a^{ik+p}) = \sum_{i=0}^{p-1} a^{-irk} \sum_{j=0}^{t} a_j a^{ikj+pj} = \sum_{j=0}^{t} a_j a^{pj} \sum_{i=0}^{p-1} a^{k(j-r)i}.$$
Because
$$\sum_{i=0}^{p-1} a^{kmi} = \begin{cases} p, & \text{if } p \mid m, \\ 0, & \text{if } p \nmid m, \end{cases}$$
we obtain
$$A_r = \sum_{j=0}^{t} a_j a^{pj} \sum_{i=0}^{p-1} a^{k(j-r)i} = p \cdot \sum_{j=0}^{\lfloor \frac{t-r}{p} \rfloor} a_{pj+r} a^{p(pj+r)}.$$

For any divisor $d \nmid kp$ consider $U_d = \{x \in K^* \mid \text{ord}(x) = d\}$ and the polynomial
$$Q_d = \prod_{x \in U_d} (X + x).$$
Let $K_2 = \{0, 1\}$ the prime subfield of $K$ and $1 = d_1 < d_2 < \dots < d_n = kp$ the divisors of $kp$. As $Q_1 = X + 1 \in K_2[X]$ and
$$Q_{d_i} = (X^{d_i} + 1) \cdot \left( \prod_{d_j \mid d_i, d_j \neq d_i} Q_{d_j} \right)^{-1} \in K_2[X],$$
we have that all polynomials $Q_{d_i} \in K_2[X]$ have all coefficients $0$ or $1$.
But then $P = (X^{kp} + 1) \cdot Q_k^{-1} \in K_2[X]$ has also this property, moreover at least $p$ being non-zero, concluding the proof.