Generate one-time-pad with pre-shared key and public randomness I know very little about cryptography, so I apologize if the question does not make sense.

Let's say we design a function $F(P, X) \rightarrow K$ that takes in a pre-shared private key $P$ and a publicly available source of randomness (say a server that generates true random numbers) $X$. The output $K$ of this function is a new key that we use as a one-time-pad to encrypt and decrypt a message. 

The sender pre-share the key $P$, the function $F$ and the source of randomness to the receiver before their communication. For every new message $M$, the sender collects the random information $X$, generates a one-time-pad $K$ using the function $F$ and encrypts the message with $K$. The receiver generates the same $K$ and decrypts the message. 

My questions are


This scheme is not information theoretic secure, right? because AFAIK, the one-time-pad cannot be reused in any way, but in this case, we are reusing $P$ for every message, which can be considered as part of the one-time-pad $K$
Is this scheme viable? Is it more secure than simply encrypting/decrypting with the pre-shared key $P$?
You can relate your construction to the approach of using a pesudo-random function (PRF) in order to extend the key-length on the one-time-pad construction (in fact, it is more typicall to use a pseudo-random generator (PRG), but since this can be obtained from a PRF, and since your construction resembles more a PRF as well, I prefer to present it like this).

A PRF is a family of functions that resemble random functions, in a very precise way.
We say that a family of functions $\{F_s\}_{s\in \{0,1\}^\ell}$ is pseudo-random, where $F_s:\{0,1\}^n\to\{0,1\}^m$, if a function sampled at random from this set (which is done by sampling a random $s\in\{0,1\}^\ell$) is indistinguishable from a totally random function $f:\{0,1\}^n\to\{0,1\}^m$ (i.e. one were the image of each $x\in\{0,1\}^n$ is a uniformly random string in $\{0,1\}^m$).
If you have such a family of functions then you can build an "extension" of the OTP by simply using the pre-shared secret key $s$ as a seed to get a random function $F_s$, and using $F_s(0)\|F_s(1)\|\cdots\|F_s(N)$ as a (much longer) OTP key.

Now, if you think about it, when you consider your function $F(P,X)\to K$ you're sort of using this idea: you are using some shared secret randomness to expand it to a longer, perhaps not totally uniform key.
So, to answer your questions, your scheme is not information-theoretic secure but not really because you're reusing the key, but because the PRF is not really information-theoretic secure.

Remarks

Notice that, in particular, the source of randomness you call $P$ doesn't even need to be random! It only needs to ensure that the same value of $P$ is not reused among several different encryptions, and this can be achieved via a counter as I illustrate above. There is some randomness involved of course, but this is coming from the shared key.