Output voltage of an amplifier circuit
This diagram explains everything:

 

Thanks to the op-amp "action" we have \$ V_+ = V \scriptstyle -\$ 

And the voltage at output (voltage drop across\$ R_2\$) is equal to

$$ V_{out} = I_L \cdot R_S \cdot \frac{R_2}{R_1} \cdot \frac{ β + 1}{\beta} \approx  I_L R_S \cdot \frac{R_2}{R_1} $$

And here you have the alternative version with a PNP transistor.



$$ V_{out} = I_L \cdot R_S \cdot \frac{R_B}{R_A} \cdot \frac{β}{\beta + 1} \approx  I_L R_S \cdot \frac{R_B}{R_A} $$

And in this circuit (with PNP) the op-amp is supplied from the same voltage as the main circuit. And the opamp \$V_{oh}\$ (positive saturation voltage) is large then 0.6V. So, we are forced to add a Zener diode in series with the transistor base to lower the op-amp output voltage.