Extremely lost and confused on how to apply law of total variance on problem I am a bit confused on the concept of conditioning and finding the new distribution of conditioned random variables.

An example, 

Suppose that there is a new carnival in town and that citizens will begin to line up outside at $10:00$ am. Further suppose that the arrivals are independent, and that the number that arrive in any 1 minute interval is Poisson distributed with $\lambda=3$.

Now suppose the owners will check on the line $T$ minutes after people begin to show up, where $T$ is binomial distributed with $n=20$ and $p=0.8$, then find the expected number of citizens in the line when it is checked and the standard deviation.

What I tried:

I know the law of double expectation and variance

the first is that $E[P]=E[E(P\mid T)]$

and $Var(P)=E[Var(P\mid T)]+Var[E(P\mid T)]$ which I could find standard deviation from.

I know that $E(T)=np=16$ and $E(P)=\lambda=3 $

If I had to give an answer without being confident, I would say that the expected number of customers would be 48

So I know that $E(C|T)$  ( I now think it is $3T$) should be a random variable, how do I find its distribution? What about in general to find the distribution of conditional expectation?

I am having some trouble on the second part and this is why, to use the formula I listed for variance above I want to calculated $Var(P|T)$ , I thought this would be equal to $E(P^2|T)-(E(P|T))^{2}$ which would give me $9T-9T^{2}$, which has an expected value of $-2188.8$, adding this to the variance of the expected value $( 28.8)$ still leaves a very negative number which I know cannot be the case. so I must have a mistake somewhere. Here I also used that we know the variance of binomial to be npq , which in this case is =3.2 and used this to solve for $E(M^2)$ I have tried over and over now but cant figure out where and why I am not getting an answer that makes sense.

Still confused, is really no one willing to help clarify this at all?? I am also confused in one of the answers how he says that the r.v is poisson

Thanks
You should have $\operatorname{E}(P\mid T) = \lambda T.$

Remember that $\lambda$ is in persons-per-minute and $T$ is in minutes, so when you multiply them you get persons.

So $\operatorname{E}(\operatorname{E}(P\mid T)) = \operatorname{E}(\lambda T) = \lambda\operatorname{E}(T) = \cdots$ etc.

And $\operatorname{var}(P\mid T) = \lambda T$ (since the variance of a  Poisson distribution is the same as the expected value.
You should have $\operatorname{E}(P\mid T) = \lambda T.$

Remember that $\lambda$ is in persons-per-minute and $T$ is in minutes, so when you multiply them you get persons.

So $\operatorname{E}(\operatorname{E}(P\mid T)) = \operatorname{E}(\lambda T) = \lambda\operatorname{E}(T) = \cdots$ etc.

And $\operatorname{var}(P\mid T) = \lambda T$ (since the variance of a  Poisson distribution is the same as the expected value.