Least Squares Filter Design: Deriving the Objective Function
$$
J(h) = \int_{R^+UR-}|e^Th-F_d|^2d\omega\tag{1}\\
= \int(e^Th-F_d)^H(e^Th-F_d)d\omega\\
= \int((e^Th)^H(e^Th) + F_d^HF_d -(e^Th)^HF_d - F_d^He^Th)d\omega\\
= h^H(\int(e^T)^He^Td\omega) h + \int |F_d|^2d\omega -\int (2 Re\{(e^Th)^HF_d\})d\omega 
$$
The second term in above integral is the integral of $L_2$ norm of $F_d$ in the region $R^+ U R^-$. Since $F_d(\omega)$ is symmetric about $\omega = 0$, this integral will be $2 \times \int_{R^+}|F_d(\omega)|^2d\omega$ which is $c$ in your questions.

The first term, the argument inside integral is the outer product of column vector $(e^H)^T$ with row vector $e^T$. Again, this will conjugate of each for the regions $R^+$ and $R^-$ (shown later in the appendix). So integral over both regions will result in the real value of integral $e^H e$.  Also, since $H$ is real, $h^H=h^T$. So the first term can be rewritten as (for simplicity dropping $T$ notation as it is assumed to be a column vector)
$$
h^T(\int_{R^+}2Re\{ ee^H\}d\omega )h\\
= h^TQh
$$ 

The third term is the result of sum of 2 conjugate terms. Again this is symmetric about $\omega=0$. Also, since $H$ is real, $h^H=h^T$. Taking out $2h^T$ from the integral we have 
$$
2h^T\int_{R+}2Re\{F_de^H \}d\omega = 2h^Tp
$$
Therefore, summing and rearranging the 3 terms,
$$
J(h) = h^TQh - 2h^Tp + c
$$

Appendix: Showing that $ee^H$ is conjugate of each other in $R^+$ and $R^-$. $E_{mn}$, element of $ee^H$ is $e^{-jm\omega}e^{-jn\omega} = e^{-j(m+n)\omega}$. In $R^-$, this element is $e^{j(m+n)\omega}$ which is $E_{mn}^*$