You can use the generating function for Schur polynomials
$$\sum_{\lambda}s_\lambda(x_1,\dots,x_m)s_\lambda(y_1,\dots,y_n)=\prod_{i=1}^m\prod_{j=1}^n\frac 1{1-x_iy_j}.$$
Take $y_1=\dots=y_n=1$, gives
$$\sum_{\lambda}s_\lambda(x_1,\dots,x_m)s_\lambda(1^n)=\prod_{i=1}^m\frac 1{(1-x_i)^n}=\sum_{k_1,\dots,k_m=0}^\infty\prod_{i=1}^m\frac{(n)_{k_i}}{k_i!}x_i^{k_i}.$$
Pick out the term of homogeneity $r$ and specialize the $x_j$,
$$\sum_{\lambda\vdash r}s_\lambda(1^m)s_\lambda(1^n)=\sum_{k_1+\dots+k_m=r}\prod_{i=1}^m\frac{(n)_{k_i}}{k_i!}=\frac{(nm)_r}{r!}$$
by the multinomial theorem.
This gives the generating function
$$\sum_{\lambda}s_\lambda(1^m)s_\lambda(1^n)t^r=\frac 1{(1-t)^{mn}}.$$
It seems you take $m=n=\kappa$ and $t=h^2$. You have
$$\log \sum_{\lambda}s_\lambda(1^\kappa)^2h^{2r}=\log \frac 1{(1-h^2)^{\kappa^2}}=\kappa^2\log \frac 1{1-h^2}.$$

Note that $s_\lambda(1^n)$ are the dimensions of the irreducible representations of $\mathrm{GL}(n,\mathbb C)$, so you can probably also give a representation-theoretic proof.