How come black body have more emissivity and more absorbtivity(a) at same time? We have two definitions to look at
Absorbtivity(a): the ratio of absorbed energy and incident energy on a body
$a_{BlackBody} = 1$

so if i have a tourch light that gives red light, in a dark room I point this tourch light at this black body, then according to above definition, I should see black color
Good absorbers are good emitters from kirchoffs law, and also stefans law
$$\frac{d\theta}{dt} = \sigma AeT^4$$
Now according to that statement, if we reconduct same experiment, when I point my red torch light towards black body it should now emmit red color

Aren't both definitions contradictory to each other or did I misinterpret something 
 A: The red light, with power $P$, will be absorbed. The blackbody will be heated to equilibrium when it's temperature $T$ satisfies:
$$ \sigma T^4 A = P $$
Thus, it all depends on $P$. If $P$ is a 1 TW red laser and $A$ is a pellet, it could get to really hot...mega kelvins. If it's a 10 W light, and $A$ is larger, the radiation will peak in microwave or mm.
A: You are mixing up two concepts, namely the emission of light as a consequence of temperature and the reflection of incident light.
When you look at objects in daylight, their apparent colour is caused by the fact that they absorb some of the incident daylight and reflect the rest. It is the combination of the frequencies of the reflected portion that determines their colour. A matt black object will absorb most of the incident light and reflect little of it, which is why it seems very dark.
Light emitted by a body as a result of its temperature is caused by a different mechanism, namely internal thermal agitation.
If you have a piece of ebony and, say, a piece of birchwood and you look at them both in daylight, the ebony will appear very dark and the birch very light, the difference being that one reflects less of the incident daylight than the other. If you put both pieces in a fire so that they burn, they will both appear orange/red, because that light is being generated internally, rather than being reflected daylight.
A: The light that the black body emits is defined by the temperature of that body. When you shine red torchlight and the body is cold, it will still look black, but as it heats up, it may start glowing red. It still will formally be called a "black body" though, since the term is reserved for a body that absorbs all the light that fall onto it.
