Right triangle where the perimeter = area*k

We have the following relations: $$P=a+b+c,\quad 2A=ab,\quad a^2+b^2=c^2$$

In fact, there is one more that is implicit: the triangle inequality $a+b>c$ (which follows from $a^2+b^2=c^2$). We wish to relate the first and second relations, and so after staring at the problem for a while we realize the correct course of action is to multiply $P$ by $a+b-c$ (the latter guaranteed to be positive by the triangle inequality). This gives:

$$\begin{align}P(a+b-c)&=(a+b+c)(a+b-c)=(a+b)^2-c^2\\ &=a^2+b^2+2ab-c^2=2ab=4A \end{align}$$

Hence in general for a right triangle we have $k=A/P=(a+b-c)/4$, so the question boils down to when the difference between the sum of the lengths of the legs and the length of the hypothenuse divides or is divisible by $4$, for a right triangle with integer side-lengths.

Here we actually need to generate pythagorean triples $(a,b,c)$ such that $a,b,c$ are integers and $a^2+b^2=c^2$; the wikipedia page here gives a good picture of Euclid's formula that I recommend thinking through. Euclid's formula is the following:

$$a=m^2-n^2,\quad b=2mn\quad c=m^2+n^2$$ for integers $m>n$.

Then the quantity $a+b-c$ is $m^2-n^2+2mn-m^2-n^2=2n(m-n)$, so $A/P=k=n(m-n)/2$. The options are now that either $2$ divides $n$ or $2$ divides $m-n$, so we either have $n=2d$ and $m=k/d+2d$ or $n=d$ and $m=2(k/d)+d$ for divisors $d$ of $k$.

(to get $P/A=k$ we need $2n(m-n)$ to divide $4$, so $n(m-n)$ must divide $2$, thus $n=1$, $m=2$, $n=1$, $m=3$, and $n=2$, $m=3$ are the only solutions, giving us $(3,4,5)$, $(8,6,10)$, $(5,12,13)$ with $k=2,1,2$ being the only possibilities).

There is a method of generating all of the Pythagorean triples, that goes back in principle at least to the time of Euclid. You may find the details in this Wikipedia article interesting.

As a brief summary, apart from the order that the two legs are listed, all Pythagorean triples $(x,y,z)$ have the shape $$x=\lambda(m^2-n^2),\qquad y=\lambda(2mn), \qquad z=\lambda(m^2+n^2),\qquad (\ast)$$ where $m$, $n$, and $\lambda$ are positive integers.

For a triangle as described in $(\ast)$, we have $P=\lambda(2m^2+2mn)$ and $A=\lambda^2mn(m^2-n^2)$. So your equations (i) $P=kA$ and (ii) $A=kP$ can be rewritten as

(i) $\lambda(2m^2+2mn)=k\lambda^2mn(m^2-n^2)$ and

(ii) $k\lambda(2m^2+2mn)=\lambda^2mn(m^2-n^2)$.

Equation (i) quickly collapses. Cancellation gives $k\lambda n(m-n)=2$, and the list of solutions is short. There are two possibilities, $k=2$ and $k=1$. If $k=2$, then $\lambda=1$ and $m=2$, $n=1$, which gives your example $(3,4,5)$. If $k=1$, then $\lambda=2$, $m=2$, $n=1$ or $\lambda=1$, $m=3$, $n=1$, or $\lambda=1$, $m=3$, $n=2$. We obtain the examples $(6,8,10)$ and $(5,12, 13)$. That's all.

Equation (ii) is more interesting. Cancellation gives $$\lambda n(m-n)=2k. \qquad(\ast\ast)$$ For any $k$, we can find a solution, for example by putting $\lambda=1$, $n=2k$, and $m=2k+1$. Or, less interestingly, we can put $\lambda=k$, $m=2$, $n=1$, which gives a scaled version of the familiar $3$-$4$-$5$ triangle.

For given $k$, we can find an explicit way to generate all solutions of $(\ast\ast)$, and to obtain a count of them. For any fixed $k$, there are only finitely many solutions. Roughly speaking, if $k$ has many divisors then there are many solutions, since we generate the solutions by looking at factors of $2k$.