$\newcommand\ka\kappa$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)][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 $\Big\{\dfrac{2c(r_p)}{\ka(p)}\colon0<p<1/6\Big\}$:

[![enter image description here][2]][2]


  [1]: https://arxiv.org/pdf/1405.4496.pdf
  [2]: https://i.sstatic.net/24l0x.png