I think I should highlight the point made about biasing, as I don't think everyone has picked up on it. In this case, biasing the DC point at half the supply voltage is undesirable. The output should be biased at half the maximum voltage swing of the amplifier output. If the maximum positive output swing of the amplifier is 6.6V (1.5V loss for the op-amp and 0.9V for the transistor, as emitter followers drop a lot more than 0.7V, at high currents) and the negative is 0V, then bias point needs to be half way between these voltages, which is 3.3V.
Using a pull down resistor will reduce the power dissipated by the speaker, but also efficiency of the circuit. If the resistor is the same value as the speaker's impedance, then the maximum power delivered to the speaker will be reduced by a factor of four, because it will drop half the voltage as P = V2/R. Using a lower value resistor will drastically increase the static power dissipation and using a higher value will reduce the maximum power output even more. Replacing the resistor with a constant current sink, but it would require an additional transistor and it will be cheaper to just use a class AB push-pull topology. Exercise for the original poster: work out the optimal current for the sink, given a voltage swing of 6V peak-to-peak (the current sink will prevent the output voltage reaching 0V) and an 8Ω speaker.