Is this extension of Goldbach's conjecture obviously false?

First of all we will assume that the Goldbach's conjecture is true, so given an integer $n=kp$ with $k\geq 4$ ($k=3,2$ are easy), we will prove that $n$ is the sum of $p$ primes.

If $p=2,$ it's the Goldbach's conjecture and it's true, so we can assume that $p>2$.

Case 1 If $n$ is even, let's write $$n=\underbrace{2+\cdots+2}_{p-2}+(n-2(p-2)),$$ and because $(n-2(p-2))=(k-2)p+4\geq4$ and it's even, then it can expressed as the sum of two primes. Hence, $n$ is the sum of $p$ primes.

Case 2 If $n$ is odd, let's write $$n=3+\underbrace{2+\cdots+2}_{p-3}+(n-2(p-3)-3),$$ and because $(n-2(p-3)-3)=(k-2)p+3\geq4$ and it's even, then it can expressed as the sum of two primes. Hence, $n$ is the sum of $p$ primes.

So we proved the following statement: $$\text{Goldbach's conjecture} \implies \text{Extension of Goldbach's conjecture}$$

But Goldbach's conjecture is a particular case of the extension of Goldbach's conjecture (when $p=2$), hence: $$\text{Goldbach's conjecture} \iff \text{Extension of Goldbach's conjecture}$$

Conclusion The extension of Goldbach's conjecture as you define it is equivalent to the Goldbach's conjecture itself, and hence no one will ever find a counterexample or prove it unless (s)he solves Goldbach's conjecture.

Suppose Goldbach is true. Let n be your candidate number. Then either n is even or odd. If even, then it is the sum of 2 primes. If odd then n - 3 is even so n is the sum of 3 primes. In no case would you need more than 3 primes.