Let us evolve the Universe in both arguments to a de Sitter end-point so the 'horizon' is then the future cosmic event horizon in both cases. What is then clearly the same in both arguments is that the entropy is:
\begin{equation}\tag{A}
\ S=S_{Bulk}=S_{dS}\sim \frac{E}{T_{dS}}
\end{equation}
Also, $T_{dS}\approx 2.4\times10^{-30}K$.
The volume of the space enclosed by the future cosmic event horizon (radius $l_\Lambda$) is simply:
\begin{equation}\tag{B}
V = \frac{4}{3} \pi{l_\Lambda^3} \sim 1.4 \times 10^{79}~\mathrm{m^3}.
\end{equation}
Or, if you feel this volume relation (B) is only valid in dS space with a horizon-based argument, you can use $V=\pi^2l_\Lambda^3$ in argument (2) below, it doesn't change the outcome.
The entropy density $s$ of the cosmic vacuum is $s=S/V$
What is not the same:
In argument (1) $E=E_H=E_{Bulk}$. This is a holographic relation. Also, $k_BS_{dS}=E_{dS}/T_{dS}$
In argument (2) $E=E_B=aT_{dS}^4V$. This a black body radiation relation. Also, $k_BS_{dS}=4E_{B}/3T_{dS}$
Lets look at (2) first. If this were right, the energy density of the cosmic vacuum could be calculated as $\rho_{\Lambda}=E_B/V=aT_{dS}^4\approx2.5\times10^{-134}\mathrm J/m^3$.
Now, the energy density of the bulk vacuum is constant over time, and from observation is known to be approx.:
\begin{equation}\tag{3}
\rho_{\Lambda} = \frac{\Lambda c^4}{8 \pi G} \sim 6.3 \times 10^{-10}~\mathrm{J/m^3}.
\end{equation}
The ratio of (3) : $\rho_{\Lambda}$ from argument (2) is $\sim10^{124}$. So, what we have with argument (2) is a version of the cosmological constant problem (CCP). In fact, we could expect this, since the CCP arises from relating the degrees of freedom to a sphere volume rather than a sphere surface area, and this is what argument (2) is really doing, trying to relate vacuum energy density to volume. Also, see these answers why this approach is a 'no'.
What about (1)? This is more likely to be correct, since as per this answer, there is a holographic-inspired approach that solves the CCP.
