In $_6^{14}\mathrm C$ decay, how is mass-energy conserved?

One needs to be very careful in doing mass-energy balances in nuclear decay reactions, especially in beta-decay (electron or positron)

The reaction as written in the OP is correct, and is exactly as it is normally written but is slightly misleading (not the fault of the OP!).

Consider an individual carbon-14 atom on the LHS of the reaction. It consists of 6 protons, 8 neutrons, and six orbital electrons. The orbital electrons are not involved in the nuclear reaction, and are usually ignored.

The major product of the decay of this carbon-14 atom is an atom of nitrogen-14. But this atom still has the same six orbital electrons as the parent carbon -14.

However, the atomic masses are tabulated for the whole neutral atom. So the mass used for the carbon-14 atom is correct, but the mass used for the daughter nitrogen-14 atom is actually too large by the mass of the seventh orbital electron found in a neutral atom. To emphasize, the actual reaction product has six orbital electrons, while the mass used is for a seven-orbital-electron nitrogen-14 atom.

So, one needs to subtract an electron from the tabulated nitrogen-14 atom, and then add back in the mass of the actual beta-particle produced in the reaction.

This is the same as just using the tabulated atomic mass values, and not including the beta-particle mass.

The atomic masses include both the nuclei and the electrons for a neutral atom. Nitrogen has one more electron than Carbon.