Confusion in electric field inside a conductor Image I'm referring to:



In the figure both shells are conductors.
Let's say I have to find the electric field at a point P as shown in the figure. I thought, since this point lies inside a conductor, the Electric field at this point has to be $0$ no matter what. But I was taught that there will be no electric field at P due to the outer shell but there will be a electric field at point P due to the inner shell, (because it has some charge).
I'm confused now, beacuse earlier I was told the electric field inside a conductor is $0$, and that point clearly lies inside the conductor, where am I going wrong? Won't there be induction of $-Q_{1}$ on the inner surface of the outer shell, in such a way that the electric field inside the conductor becomes $0$?
Also, by the term inside the conductor mean in the bulk or in the material of the conductor or something else?
Please help me out, I have studied from many sources, any no one cares to be critical about their words, hence the confusion.
Edit: For all future readers, this image is a good explaination of what is going on inside such situations.

 A: As Matteo points out, it is better to use Gauss's law to avoid any confusion.

...earlier I was told the electric field inside a conductor is $0$

This is true only when put in conjunction with another statement you may have come across:

If the conductor is isolated and carries a charge, the charge resides on it's surface.

Hence, if we apply Gauss's law to a gaussian sphere inside the conductor, as $q_\text{in} = 0$ it follows that $E = 0$  ($\vec E$ and $d\vec A$ are not perpendicular)
Now, in the question, if you draw a gaussian sphere that is concentric with either shell and passes through P, you will find that $q_\text{in} = Q_1$ is not zero, and hence, $|E| > 0$
Hope this helps.
A: The most reliable way to approach these problems is to use fundamental laws of electromagnetism, in particular Gauss' law and linearity of the electric field. A thorough application of these laws can answer all of your questions.
Gauss law
The electric field flux through a closed surface $\Phi_E$ is given by the charge inside that surface $Q_{int}$ divided by $\varepsilon_0$:
$$\Phi_E = \frac{Q_{int}}{\varepsilon_0 }. $$
Linearity of electric field
The total electric field at a point is the vector sum of all the electric field produced by the single sources. If there are two sources labeled $1$ and $2$, the field at a point $\vec{E}(P)$ is
$$ \vec{E}(P) = \vec{E}_1(P) + \vec{E}_2(P). $$
Let's now use these tools to solve your problem: at point P there are in principle two contributions to the field, one coming from the inner shell, and one coming from the outer shell. The former has modulus $Q_1/4\pi\varepsilon_0r^2$ and the latter vanishes, as a consequence of Gauss' law (exercise). Bottom line: there is an electric field at P, only coming from the inner shell charges.
Now let's comment your statements one by one:

I thought, since this point lies inside a conductor, the Electric
field at this point has to be 0 no matter what.

This is a misunderstanding, since "inside a conductor" means "in the bulk" of a conductor, but here there's no bulk at all.

I was taught that there will be no electric field at P due to the
outer shell but there will be a electric field at point P due to the
inner shell

This is correct, as discussed above.
EDIT
while writing this answer, I was unaware of most of the comments and other answers, so there might be some redundancy.
