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).
Best Answer
Since its a right-angled triangle, follow these 4 steps to get the solution:
Step-1: Angle is given use the Trigonometric relation of Tangent:
Step-2: Triangle area formula i.e.
Now you can substitute one unknown variable from eq(1) to the eq(2) or vice-versa to obtain both the Base and Height of the triangle.
Step-3: Use the Pythagoras theorem
Step-4: Perimeter = Hypotenus + Height + Base