Let $z_1, \dotsc, z_n$ be the (not necessarily distinct) roots of $P$, so that
$$P(z) = \prod_{j=1}^n (z- z_j) .$$
Since all the coefficients of $P$ are real, it follows that if $w$ is a root of $P$, then $P( \overline{w}) = \overline{ P(w)} = 0$, so $\overline{w}$, the complex conjugate of $w$, is also a root of $P$.
Since
$$\lvert i- z_1 \rvert \cdot \lvert i - z_2 \rvert \dotsm \lvert i - z_n \rvert = \lvert P(i) \rvert < 1,$$
it follows that for some (not necessarily distinct) conjugates $z_i$ and $z_j$,
$$\lvert z_i-i \rvert \cdot \lvert z_j-i \rvert < 1.$$
Let $z_i = a+bi$ and $z_j = a-bi$, for real $a,b$. We note that
$$(a+b+1)^2 - (a+b-1)^2 = 4a + 4b .$$
Thus
$$\begin{align*} (a^2+b^2+1)^2 &= (a^2+b^2-1)^2 + 4a^2 + 4b^2 = \lvert a^2 + b^2 - 1 - 2ai \rvert ^2 + 4b^2 \\ &= \lvert (a-i)^2 - (bi)^2 \rvert^2 + 4b^2 \\ &= \bigl( \lvert a+bi - i \rvert \cdot \lvert a-bi -i \rvert \bigr)^2 + 4b^2 \\ &= \bigl( \lvert z_i - i \rvert \cdot \lvert z_j - i \rvert \bigr)^2 + 4b^2 < 1+4b^2. \end{align*}$$
Since $P(a+bi) = P(z_i) = 0$, these real numbers $a,b$ satisfy the problem's conditions. $\blacksquare$