To summarize the discussion in the comments:

There are $\pi(n)$ primes less than or equal to $n$ (this is standard notation). This means there are $\pi(n-1)$ primes less than $n$.

The Goldbach conjecture would tell us that each even number ($>2$) is the sum of two primes. Thus, if $n>3$ were odd, $n-p_{\pi(n-1)}$ is even so Goldbach tells you want you want, at least in the case where $n-p_{\pi(n-1)}>2$. If you allow $a=b$, then you are ok even in the case $n-p_{\pi(n-1)}=2$ (as you can take $a=b=1$).

If $n$ is even, however, then $n-p_{\pi(n-1)}$ is odd so you'd need one of $a,b$ to be $2$, which would force $n-p_{\pi(n-1)}-2$ to be prime, which of course it usually is not. Thus your conjecture is generally false for even numbers. Of course, if you drop the condition that one of the summands be $p_{\pi(n-1)}$ you can apply Goldbach to $n-3$.

To summarize the discussion in the comments:

For $n\in \mathbb N$ (not restricting to primes) there are $\pi(n)$ primes less than or equal to $n$ (this is standard notation). This means there are $\pi(n-1)$ primes less than $n$.

The Goldbach conjecture would tell us that each even number ($>2$) is the sum of two primes. Thus, if $n>3$ were odd, $n-p_{\pi(n-1)}$ is even so Goldbach tells you want you want, at least in the case where $n-p_{\pi(n-1)}>2$. If you allow $a=b$, then you are ok even in the case $n-p_{\pi(n-1)}=2$ (as you can take $a=b=1$).

If $n$ is even, however, then $n-p_{\pi(n-1)}$ is odd so you'd need one of $a,b$ to be $2$, which would force $n-p_{\pi(n-1)}-2$ to be prime, which of course it usually is not. Thus your conjecture is generally false for even numbers. Of course, if you drop the condition that one of the summands be $p_{\pi(n-1)}$ you can apply Goldbach to $n-3$.

To summarize the discussion in the comments:

The Goldbach conjecture would tell us that each even number ($>2$) is the sum of two primes. Thus, if $n>3$ were odd, $n-p_{n-1}$ is even so Goldbach tells you want you want, at least in the case where $n-p_{n-1}>2$. If you allow $a=b$, then you are ok even in the case $n-p_{n-1}=2$ (as you can take $a=b=1$).

If $n$ is even, however, then $n-p_$ is odd so you'd need one of $a,b$ to be $2$, which would force $n-p_{n-1}-2$ to be prime, which of course it usually is not. Thus your conjecture is generally false for even numbers. Of course, if you drop the condition that one of the summands be $p_{n-1}$ you can apply Goldbach to $n-3$.

To summarize the discussion in the comments:

There are $\pi(n)$ primes less than or equal to $n$ (this is standard notation). This means there are $\pi(n-1)$ primes less than $n$.

The Goldbach conjecture would tell us that each even number ($>2$) is the sum of two primes. Thus, if $n>3$ were odd, $n-p_{\pi(n-1)}$ is even so Goldbach tells you want you want, at least in the case where $n-p_{\pi(n-1)}>2$. If you allow $a=b$, then you are ok even in the case $n-p_{\pi(n-1)}=2$ (as you can take $a=b=1$).

If $n$ is even, however, then $n-p_{\pi(n-1)}$ is odd so you'd need one of $a,b$ to be $2$, which would force $n-p_{\pi(n-1)}-2$ to be prime, which of course it usually is not. Thus your conjecture is generally false for even numbers. Of course, if you drop the condition that one of the summands be $p_{\pi(n-1)}$ you can apply Goldbach to $n-3$.