What would happen if all dark energy (cosmological constant) instantly converted to radiation due to vacuum metastability event, while otherwise vacuum properties remain the same?
Dark energy density is $10^{-27} kg/m^3$. How will it affect Earth and the Sun?
Convert mass density to energy density, to get $9 \times 10^{-11}\,\mathrm{J/m^3}$ which isn't very much. After all, an adult human is on order of one eighth to one quarter of a cubic meter.
So, assuming a blackbody spectrum so that the radiation is essentially all low energy and does not ionize the immediate temperature rise will be trivial. The planet itself will be similarly subject to a trivial increase in total energy (it's a lot of energy for the whole Earth, but it is spread out over a very large volume and it a pittance compared to the energy already present).
But all of space that is not filled with matter will be radiating in all direction. We need to know how hot that radiation field is.
The total energy density of a blackbody field is $$ u = \frac{4 \sigma T^4}{c} \;,$$ where $\sigma = 5.67 \times 10^{-8} W m^{-2} K^{-4}$. So, solving for temperature we get $$ T = \left( \frac{u c}{4 \sigma} \right)^{0.25} = 19 \,\mathrm{K} \;,$$
Overall result on earth: pretty boring, except that radio astronomers will be vexed.
Very cold objects in space will tend to warm up, but even Kuiper belt objects are a bit warmer than this (circa $50 \,\mathrm{K}$) as far as we know.
Basically a big "Meh!".
If you want it to be exciting make a non-thermal assumption for the radiation field.
