I think your cup-product is always zero independent of Leopoldt. Consider $\mathbb{Z}_p$-coefficients instead (enough by Neukirch-Schmidt-Wingberg, 2.3.10). If $\chi=1=\chi^{-1}$, then both $H^1$ degenerate to $H^1(\tilde{K}/K,\mathbb{Z}_p)$ where $\tilde{K}$ is the compositum of all $\mathbb{Z}_p$ extensions, because cocyles are Hom's. In particular, your cup-product factors through
$$
H^1(\tilde{K}/K,\mathbb{Z}_p)\times H^1(\tilde{K}/K,\mathbb{Z}_p)\xrightarrow{\cup}H^2(\tilde{K}/K,\mathbb{Z}_p)
$$
but the $H^2$ is trivial because a free $p$-group has $p$-cohomological dimension $1$ (and NSW, 2.3.5 tells you that cohomology with $\mathbb{Z}_p$-coefficients is the projective limit of those with $\mathbb{Z}/p^n$-coefficients). So, your cup-product is zero if $\chi=1$.

In the general case, by NSW, 1.5.3, cup-product is compatible with inflation and restriction, and by Inf-Res you are reduced to the case
$$
H^1(\Delta,\mathbb{Z}_p(\chi))\times H^1(\Delta,\mathbb{Z}_p(\chi^{-1}))\xrightarrow{\cup} H^2(\Delta,\mathbb{Z}_p)
$$
where $\Delta=G_K/\mathrm{Ker}(\chi)$ and now both $H^1$ vanish.