Let's try to solve this problem starting from where you're at.
If the question's asking for different ordered doubles such that $x+y=50$ then there're $51$ different ordered pairs. So, the equation that $x+y=n$ can be satisfied by $n+1$ different ordered pairs $(x,y)$:
$$(0,n),(1,n-1),\ldots ,(n-1,1),(n,0)$$
For your case, asking for different ordered triples, we can partition the solution of $a+b+c=50$ into disjoint cases in which $c=0,c=1,\ldots ,c=49,c=50$.
So, if $c=0$, then $a+b=50\implies $ there are $50+1=51$ possible answers for $(a,b,0)$
Now, for $c=1\implies a+b=49\implies $ there are $49+1=50$ possible answers for $(a,b,1)$
$$\vdots$$
So, the general trend is that for each $c=m \implies$ that there are $50-m+1$ ways to choose $a$ and $b$ for that specific $c$.
$\therefore$ There are $51+50+\ldots +3+2+1=\frac{51\times 52}{2}=1326$ different ordered triples.
I'm using "ordered" (as you clarify in the comments) to mean that the order of $(p,q)$ is important and makes a distinct pair, unless $p=q$.
Let's work up to this by solving for $4=2^2$ then $36=2^2\cdot 3^2$.
So the number of $(p,q)$ for $\text{lcm}(p,q)=4$ relies on having one of $p,q=4$, and the other can be $\in \{1,2,4\}$ - that is, $2+1=3$ options from the exponent going from $0$ to $2$. This gives us $3\times 2 = 6$ solutions (the ordering giving the $\times 2$), but then we remove the double counted $(4,4)$ giving $5$.
Now for $36$: We can solve independently as above for the powers of $2$ and $3$ but the only duplicate will be $(36,36)$ since no other repeated number will give lcm of $36$. So we have $6\times6-1 = 35$ solutions. (If we wanted "unordered" $(p,q)$, meaning $(18,4)\equiv (4,18)$, we would divide by $2$ rather than subtract $1$.)
Now it's clear that the number of solutions for $8100=2^2\cdot3^4\cdot5^2$ is $6\times 10\times 6 - 1 = 359$ options for $(p,q)$.
And, for the follow-on part of the question, $K=359$ is a prime number.
Best Answer
You need to find the number of cases with $a \lt b \lt c$, $a=b\lt c$, $a \lt b =c$ and $a=b=c$, though you can join the second and third together into a set called "two equal", and we can call the first "none equal" and the last "three equal".
"Three equal" is easy: you want solutions to $3p=9$ and $3s=9$ and there only is $1 \times 1 = 1$ "three equal" solution.
"Two equal" requires the number of solutions to $2p+r=9$ and $2s+u=9$. This is $5 \times 5 = 25$. These will each appear once either as $(2^p5^s,2^p5^s,2^r5^u)$ or as $(2^r5^u,2^p5^s,2^p5^s)$ so we do not need to adjust for duplication. But $1$ is the "three equal" solution, leaving $25-1=24$ "two equal" solutions.
"None equal" is a little harder. You have the $55\times 55$ you found. But you need to subtract three times the $24$ "two equal" solutions [e.g. if you have a solution $(a,a,c)$ then it will appear also as $(a,c,a)$ and $(c,a,a)$] and one times the "three equal" solution. Even then five-sixths of these "none equal" solutions are in the wrong order. So you want $\frac{55\times 55-3\times 24 -1 \times 1}{6}=492$ "none equal" solutions.
And then your answer is $1+24+492=517$ ordered solutions as lnwvr has found.