Not sure how to show $p_{n+1} \leq 4p_1...p_n - 1$, where $p$ are primes of the form $4n + 3$.
Only way that leaps out to me right now is thinking about it intuitively, the right side being a larger number than the next prime but I'm not sure how to show that.
On
We always have $$ p_{n+1}\le p_1p_2\cdots p_n+1\le 4p_1p_2\cdots p_n-1. $$
The second estimate is clear because of $3p_1p_2\cdots p_n\ge 2$. The first estimate is shown here:
$(n+1)th$ prime $p_{n+1}$ less than or equal to $p_1p_2\dots p_n+1$
So you do not need the assumption that the primes are of the form $4n+3$.
Problem statement: I think the intended problem statement is the following.
Let $p_k$ denote the $k$-th prime congruent to $-1$ mod $4$, i.e. the $k$-th prime of the form $4n + 3$ for some $n$. Show that for all $k \ge 1$, $$ p_{k+1} \le 4 p_1 p_2 \cdots p_k - 1. $$
Solution: the integer $n(k) = 4 p_1 p_2 \cdots p_k - 1$ is not divisible by $p_j$ for $1 \le j \le k$, and is congruent to $-1$ mod $4$. Now we have to use that every integer congruent to $-1$ mod $4$ must have at least one prime factor congruent to $-1$ mod $4$. Let $p$ be a prime factor of $n(k)$ that is congruent to $-1$ mod $4$. Then $p \le n(k)$ and $p$ is not equal to any of $p_1, \dots, p_k$. It follows that $p_{k+1} \le p \le n(k)$.