Without knowing a lot more about the details of the entire system, we can't say exactly what the difference will be between the AC1+AC2 case and the AC2-only case. However, there are a few things we can work out from simple energy balance.
In the following, I'm considering the air as an ideal gas with a fixed specific heat capacity at constant pressure $c_p$ in the temperature range of interest. This means that no account is made of the different relative humidity of air coming from the two systems. Since the air coming from the AC units is pretty dry anyway, this should be a fairly good approximation.
We have some mass of air (internal to the office) passing through each of the air-con systems per unit time: denote the indoor mass flow rate through AC$i$ as $\dot{m}_i$. There is also a source of heat from outside the building (through the walls, windows, doors etc.) which we denote $Q_{th}$. This isn't known based on the details in the question, but a pretty reasonable assumption is that it's proportional to the temperature difference between the outside air and the inside air. Call it $\dot{Q}_{th}=C(T_\mathrm{out}-T_\mathrm{in})$, where $C$ is some 'thermal transfer coefficient'. Now, we know that in equilibrium we must have energy out = energy in. In terms of the figure, this equates to
$$12C+3c_p\dot{m}_1-5c_p\dot{m}_2=0\:.$$
We wish to know how much lower the temperature in the office (currently 296 K) will be if we switch off the unit AC1. This is equivalent to setting the mass flow rate $\dot{m}_1$ to zero. We assume that, since the room was warmer than its set-point temperature, AC2 was already working flat out and so neither $\dot{m}_2$ nor the output temperature change. The equation for energy balance with our new internal temperature $T_{new}$ is
$$(308-T_{new})C-(T_{new}-291)\dot{m}_2=0\:.$$
Equating these two expressions (since $0=0$), we can rearrange to get
$$T_{new}=296-\frac{3c_p\dot{m}_1}{(C+c_p\dot{m}_2)}\:.$$
We know something else which allows us to eliminate one of the quantities in this expression: in the limit $C=0$, i.e. where the room is hermetically sealed, $T_{new}$ must equal the output temperature of AC2, i.e. 291 K. We can set $C=0$ in the above expression to find that $\dot{m}_1=5\dot{m}_2/3$. Plugging this in, we therefore have our final expression
$$T_{new}=296-\frac{5c_p\dot{m}_2}{(C+c_p\dot{m}_2)}\:.$$
This expression has the behaviour we're looking for; it goes to 291 K when $C$ goes to zero, and as $C$ gets more significant relative to $c_p\dot{m}_2$ it approaches 296 K.
And that, I think, is about as far as we can go without knowing in full detail the thermal properties of the office such as its insulation, surface area, material composition and everything else!
