In my second comment, try to explain it can be said that $$p_n\leq 2^{n-k(n)}p_{[k(n)]}$$, which is nothing but a generaliazition of above idea.

Let us assume that $k(n)=n-1$ for example. Then we know all prime numbers $p_1,p_2\ldots , p_{n-1}$. It is known that there exists a prime number between $m$ and $2m$ for $m\geq 2$. Thus, there exists a prime between $p_{n-1}$ and $2p_{n-1}$, which means $p_n\leq 2p_{n-1}$. That is, $p_n$ can not be arbitrarily large. By using thi idea, in my first comment I showed that "good $k(n)\leq n-1$".

@user7212389: I did not understand most of your comments, even if I would like to understand. I edited the question. By the way I did not mean that $p_i<k(n)$. I mean that $i\leq k(n)$. That is why I do not understand your sentence "You failed to specify $p_i<k(n)$ in the question."

In first comment, I showed that $k(n)\leq n-1$.

I did not understand what you mean, did you read the first comment ?

I think the reason is the following; You are claiming there are infinitely many prime number satisfying the congruence relation, which is true. But, $p_n$ is bounded by $k(n)$. More specifically, $$p_n \leq p_{[k(n)]}2^{n-k(n)}$$ Ofcourse, there are better bounds also. Thus, the question turns to be whether there is a unique solution satisfying the inequality.

If I am wrong, let me know I claimed that $k(n)\leq n-1$. Now we know all primes $p_1,...,p_{n-1}$. Then $p_n\leq 2p_{n-1}$. Now, since I know $x_1$ and $x_{n-1}$, and $p_1=2$ we know that $p_n\equiv t \ mod \ 2p_{n-1}$. Since $p_n\leq 2p_{n-1}$, $p_n=t$.

@MikkoKorhonen: Did you check by the way?

@MikkoKorhonen: I guess I fixed the argument.

@MikkoKorhonen: Thank you. Let me check that whether I can fix the proof for the cae $p$ is the smallest prime.

you are definitely right. I could not see the mistake in the proof.

Can you please explain "symmetric generating set " ?