Correct choice is (c) Voltage divider bias
The explanation is: When the transistor starts operating, temperature at the junction increases. Hence IC increases. As a result of which, Ie increases. Due to this increase in Ie, voltage drop across Re increases. This reduces the forward voltage across the emitter. Thus Ib reduces.
We know that, for any value of ICO, IC = β * Ib +(1+ β) ICO
where ICO =reverse saturation current (increases with temperature)
β = gain
Therefore, β and ICO increases and at the same time there is decrease in Ib. Hence above equation confirms that IC can be maintained within limits. Thus the circuit is more thermally stable and the operating point is more stable.