In what way are the Concorde's engines considered efficient?
You compare SFCs at different speeds. That is like comparing payloads for differently sized aircraft. SFC goes up with speed and, therefore, must be compared at the same speed. The work performed by an engine is thrust times distance, and higher speed means that the same thrust will perform more work per unit of time when the engine moves faster. The moving engine needs to slow down the airflow for combustion to take place, and then needs to accelerate the air by more than it has been slowed down to have positive thrust. Hence, SFC goes up in parallel with speed.

To have a meaningful comparison, we need to define efficiency. There are several, and two are of major importance for air-breathing aircraft engines: Thermal efficiency and propulsive efficiency.

Thermal efficiency

This describes how efficiently the chemical energy in the fuel $Q$ is converted into an impulse change of the air flowing through the engine. Formulated using the mass flow per unit of time $\dot{m}$, the impulse is $\dot{m}\cdot\dfrac{\Delta v^2}{2}$. Using $v_{\infty}$ for the incoming air speed and $v_{\infty} + \Delta v$ for the exit flow speed, the thermal efficiency is $$\eta_{therm} = \frac{\dot{m}\cdot \left((v_{\infty} + \Delta v)^2 - v_{\infty}^2\right)}{2\cdot Q}$$
To achieve good efficiency at high speed, a high $\Delta v$ is helpful. This explains why efficiency drops more over speed for high-bypass ratio engines and especially propellers.
Since the thermal energy in fuel is the same for all engines in your question, because all run on kerosene, and we can assume a similar efficiency of combustion, we can neglect $Q$ in the comparison.

Propulsive efficiency

This describes how well the conversion is performed. Using the same variables as above, propulsive efficiency is $$\eta_{prop} = \frac{v_{\infty}}{v_{\infty} + \frac{\Delta v}{2}}$$

This equation explains the better efficiency of high-bypass ratio engines and propellers at the same speed, because propulsive efficiency is proportional to the inverse of $\Delta v$.

Overall efficiency

This is the product of thermal and propulsive efficiency, and the equation is $$\eta_{total} = \frac{T\cdot v_{\infty}}{Q}$$
where $T = \dot{m}\cdot\Delta v$ denotes the thrust. Conveniently, $\Delta v$ is eliminated in the product, allowing turbojet engines like the Olympus&nbsp;593 to look much better in comparison to other engines.

Intake efficiency

This answer would be incomplete without a look at the intake of the Concorde. At cruise, it would lift the pressure of the air at the compressor face by a factor of more than six over ambient by efficiently decelerating the flow. The compressor added a compression ratio of 12, so the pressure in the combustion chamber was 80 times higher than ambient. This high pressure makes the engine so efficient, but is also needed to maintain combustion. Remember, ambient pressure in 18&nbsp;km is just 76 mbar, so the absolute pressure in the combustion chamber at cruise was only 6 bar.

The full answer would be like this: The combination of intake and Olympus&nbsp;593 at Mach&nbsp;2.02 had a very good total efficiency, and comparisons with other engines at static conditions are misleading.

The comparison of results from a test stand on the ground would yield a very different picture, however.