Finding Expectation Given Conditional Expectation

Suppose random variable $$Y_{il}$$ has the expectation:

$$E(Y_{il}) = \int E(Y_{il} | Z_i= z)P^{Z_i}(dz)$$

I am given that

$$E(Y_{il} | Z_i= z) = 1-exp(-\lambda_izB)$$

Thus we get end up with the following equation for the Expectation.

$$E(Y_{il}) = \int (1-exp(-\lambda_izB)P^{Z_i}(dz))$$

Skipping over the constants $$\lambda_i, B$$ because they are irrelevant to the question.

Suppose that $$Zi$$ ~ $$\gamma(\theta,\frac{1}{\theta})$$, and $$P^{Z_i}$$ is the probability distribution for Zi.

My Question: Because in the conditional expectation we condition on $$Z_i= z$$, Does this mean that our probability distribution also in terms of $$Z_i= z$$ or just a generic random variable $$Z_i$$?

Apologies if this is a stupid question.

You have $$E(Y)=\int E(Y|Z=z)p(Z=z)dz \text { integral rule}\\ =\int(1-e^{-\lambda Bz})p(Z=z)dz\\ =\int_{-\infty}^\infty(1-e^{-\lambda Bz})\frac{(1/\theta)^\theta}{\Gamma(\theta)}z^{\theta-1}e^{-\frac z \theta}dz$$
Indeed in your very first line you should write $$P^{Z=z}$$ not $$P^Z$$ otherwise that equality is not technically true.