$\newcommand\ka\kappa$Let us show that \begin{equation*} \ka(p)=p\quad\text{for }p\in[1/6,1/2]. \tag{1}\label{1} \end{equation*}
Indeed, suppose that $p\in[1/6,1/2]$. Write \begin{equation*} \ka(p) = \sup_{t>0}f_p(t), \end{equation*} where \begin{equation*} f_p(t):=\frac{\ln(1-2p+2p\cosh t)}{t^2}. \end{equation*} For real $t>0$, let \begin{equation*} g_p(t):=f'_p(t)\frac{t^3}2=\frac{pt\sinh t}{1-2p+2p\cosh t}-\ln(1-2p+2p\cosh t). \end{equation*} Then \begin{equation*} \begin{aligned} h_p(t)&:=g'_p(t)\frac{(1-2p+2p\cosh t)^2}p \\ &=t \cosh t-\sinh t-4 p (t+\sinh t) \sinh ^2\frac t2 \\ & \le h_{1/6}(t)=-\sum_{k\ge2}\frac{2^{2k+1}-8k}{6(2k+1)!}\,t^{2k+1}<0, \end{aligned} \end{equation*} so that $g_p(t)$ is decreasing (in $t>0$). Also, $g_p(0+)=0$. So, $g_p(t)<0$ (for $t>0$) and hence $f_p(t)$ is decreasing (in $t>0$). So, \begin{equation*} \ka(p) = f_p(0+)=p, \end{equation*} which proves \eqref{1}.
For $p\in(0,1/6]$, a quick bound on $\ka(p)$ can be obtained from the Kearns--Saul inequality -- see e.g. inequality (1.1), which states that the subgaussian constant for a centered Bernoulli random variable (r.v.) $X$ with parameter $r\in(0,1)$ is no greater than \begin{equation} c(r):=\frac{1-2r}{4\ln(1/r-1)}. \end{equation} Note that, if $p\in(0,1/4]$, $X$ is as above, $Y$ is an independent copy of $X$, and $r=r_p:=\frac12-\sqrt{\frac14-p}$, then $X-Y$ will equal your r.v. $\xi$ in distribution. It follows that for $p\in(0,1/6]$ \begin{equation} \ka(p)\le2c(r_p). \end{equation}
Here is the graph $\Big\{\dfrac{2c(r_p)}{\ka(p)}\colon0<p<1/6\Big\}$:
