For every $c \in \mathbb N$ either there exists at least one $(a,b) \in \mathbb N \times \mathbb N$ such that $a^2+b^2=c^2$ or such $(a,b)$ do not exist.
We could define counting function $C$ to mean $C(n)= \text{the number of pairs $(a,b)$ such that $a^2+b^2=c^2$}$ and $\text {$(a,b,c)$ is a primitive Pythagorean triple for $1 \leq c \leq n$}$
Now, the other counting function $P$ can be defined to mean $P(n)=\text{the number of numbers of the form $a+b+c-1$ or $a+b+c+1$}$ such that either $a+b+c-1$ or $a+b+c+1$ (or both) are primes for $1 \leq c \leq n$ and $(a,b,c)$ is a primitive Pythagorean triple
It should be noted that if both $a+b+c-1$ and $a+b+c+1$ are primes then only one of them is counted in $P(n)$.
If very very often either $a+b+c-1$ or $a+b+c+1$ (or both) are primes then it seems possible to have $$\lim_{n \to + \infty} \dfrac {P(n)}{C(n)} >0$$ but this seems to be highly improbable and would defy even the most optimistic heuristics that would opt for this result to hold.
So if $$\lim_{n \to + \infty} \dfrac {P(n)}{C(n)} =0$$ then how to prove that?
Some ideas?
I developed a formula that generates all Pythagorean triples where $GCD(A,BC)=2x-1,x\in\mathbb{N}$ ...which includes all primitives.
$$A=(2n-1)^2+2(2n-1)k\qquad B=2(2n-1)k+2k^2\qquad C=(2n-1)^2+2(2n-1)k+2k^2$$
where $n$ is a set number and $k$ is a member number or "count" within a set as seen in the sample below.
$$\begin{array}{c|c|c|c|c|c|c|} n & k=1 & k=2 & k=3 & k=4 & k=5 & k=6 \\ \hline Set_1 & 3,4,5 & 5,12,13& 7,24,25& 9,40,41& 11,60,61 & 13,84,85 \\ \hline Set_2 & 15,8,17 & 21,20,29 &27,36,45 &33,56,65 & 39,80,89 & 45,108,117 \\ \hline Set_3 & 35,12,37 & 45,28,53 &55,48,73 &65,72,97 & 75,100,125 & 85,132,157 \\ \hline Set_{4} &63,16,65 &77,36,85 &91,60,109 &105,88,137 &119,120,169 & 133,156,205 \\ \hline Set_{5} &99,20,101 &117,44,125 &135,72,153 &153,104,185 &171,140,221 & 189,180,261 \\ \hline Set_{6} &143,24,145 &165,52,173 &187,84,205 &209,120,241 &231,160,281 & 253,204,325 \\ \hline \end{array}$$
Only $Set_1$ contains only primitives (note $C-B=(2n-1)^2=1$). In all other sets, a multiple is generated any time any $k$ is divisible by a factor of $(2n-1).\quad$ $C$ can be prime in any of the sets but only in $Set_1$ can $A$ be prime because $A=(2n-1)^2+2(2n-1)k\quad=\quad (2n-1)(2n-1+k)\quad$ and if $n>1$ then $A$ is composite.
In the first $15$ members of $Set_1$ where $A+B+C\pm1\lt 1000$, we have $5$ prime pairs and $8$ single primes and, like primes anywhere, they seem to get scarcer with altitude. Here the triples and counts ()[].
$$ (3,4,5)[2]\quad (5,12,13)[2]\quad (7,24,25)[1]\quad (9,40,41)[1]\quad (11,60,61)[2]\quad (13,84,85)[1]\quad (15,112,113)[1]\quad (17,144,145)[1]\quad (19,180,181)[2]\quad (21,220,221)[0]\quad (23,264,265)[1]\quad (25,312,313)[1]\quad (27,364,365)[0]\quad (29,420,421)[2]\quad (31,480,481)[1]\quad $$
We have infinite sets and only $Set_1$ can contain double primes. Granted the value of $C$ grows faster in the other sets for a given $k$ but, for given limits on the value of $C$, we still have a growing finite number of double primes $Set_1$ divided a faster growing number of single primes only in the other sets. Therefore, if you can formalize the logic, I think you may be able to prove your conjecture that the limit of $\frac{P}{C}=0$.
Update:$\quad$ I was concerned about multiples outnumbering primitives as the values of $(2n-1)$ composites got ever larger and had more and more factors that could divide $k$ but I remembered something which allows us to ignore $(2n-1)$ composites completely.
$(2n-1)$ generates all primes greater than $2$ so there are an infinite number of sets where $(2n-1)$ is prime. To avoid generating multiples, if $(2n-1)$ is prime, we need only modify the formula so that it repeatedly generates $\big((2n-1)-1\big)$ triples, then skips one.
\begin{align*} &A=(2n-1)^2+&2(2n-1)\bigg(k+\bigg\lfloor\frac{(k-1)}{(2n-2)}\bigg\rfloor\bigg)&\qquad\\ &B=&2(2n-1)\bigg(k+\bigg\lfloor\frac{(k-1)}{(2n-2)}\bigg\rfloor\bigg)&\qquad+2\bigg(k+\bigg\lfloor\frac{(k-1)}{(2n-2)}\bigg\rfloor\bigg)^2\\ &C=(2n-1)^2+&2(2n-1)\bigg(k+\bigg\lfloor\frac{(k-1)}{(2n-2)}\bigg\rfloor\bigg)&\qquad+2\bigg(k+\bigg\lfloor\frac{(k-1)}{(2n-2)}\bigg\rfloor\bigg)^2 \end{align*}
For any $Set_n, (n>1\land (2n-1) \text{ is prime})$, this formula will skip multiples and generate a set containing only primitives. Since there are infinite triples in each set, each will contain an infinite number of prime $C$-values. Also, there are infinite sets where $(2n-1)$ is prime so we effectively have $$\frac{\infty}{\infty^2}$$.
This makes no sense mathematically because $\infty^2=\infty$ but we are dealing with finite counts and it is easy to see that the limit approaches this $philosophical$ $\frac{x}{x^2}=0$.
Two more trivia that may not help but are interesting. $\mathbf{(1)}$ All primes greater than the sub-primes $2\&3$ take the form of $6x\pm1$ while all C-values take the form of $4n+1$. This means that $C$ will never have the values of $7,11,19,23,31,43,...$ but there may be a prime whenever the two can generate the same number. $\mathbf{(2)}$ There can be multiple triples for some composite $C$-values but not if $C$ is prime. This means that a prime $C$-value in any set will be unique among all sets.
Good luck in your proof.