Let $\chi$ be the nontrivial Dirichlet character modulo $4$, and let $L(s,\chi)$ be the Dirichlet L-function
$$L(s,\chi) = 1 - \frac{1}{3^s} + \frac{1}{5^s} - \cdots = \sum\limits_{n=1}^{\infty}\frac{\chi(n)}{n^s} = \prod\limits_p (1- \frac{\chi(p)}{p^s})^{-1}$$
which converges absolutely for $\Re s > 1$. This L-function is in fact analytic on $\Re s > 0$ and satisfies $L(1,\chi) = \frac{\pi}{4}$. Now for $\sigma > 1$,
$$\log L(\sigma,\chi) = \sum\limits_p \frac{\chi(p)}{p^{\sigma}} + G(\sigma)$$
where $G(\sigma) = \sum\limits_{n=2}^{\infty} \sum\limits_p \frac{\chi(p)^n}{np^{n\sigma}}$ converges absolutely for $\sigma > \frac{1}{2}$.
If I have understood all this correctly, then we should have
$$\sum\limits_p \frac{\chi(p)}{p} = \lim\limits_{\sigma \to 1^+} \sum\limits_p \frac{\chi(p)}{p^{\sigma}} = \log L(1,\chi) + G(1) = \log \frac{\pi}{4} + G(1) < \infty$$
But the series $\sum\limits_p \frac{\chi(p)}{p}$ is conditionally convergent, since $\sum\limits_p \frac{1}{p}$ diverges. So how can we say that the sum $\sum\limits_p \frac{\chi(p)}{p}$ is an unambiguous finite real number? Conditionally convergent series can be rearranged to have an arbitrary limit.
I think the problem lies on the fact that by saying $$L(1,\chi)=\frac{\pi}{4}<\infty$$ you are already using the formula $$\sum_{n=1}^{\infty}\frac{x^n}{n}=-\log(1-x)$$ to define the value $L(1,\chi)$ for $\chi\neq 1$, even if this series only converges conditionally for $\vert x\vert=1$ and $x\neq 1$.
To see this in general, suppose $\chi$ is a non-trivial primitive Dirichlet character mod N, you can define the Gauss sum to be $$\tau(\chi)=\sum_{n \mod N}\chi(n)\psi(n)$$ where $\psi(n)=\exp(2\pi i n/N)$. Using this one can show that the equality holds:
$$\sum_{n\mod N}\chi(n)\psi(nm)=\overline{\chi(m)}\tau(\chi).$$
Since $\chi(-1)\overline{\tau(\chi)}=\tau(\overline{\chi})$, together with the fact that $\tau(\chi)\overline{\tau(\chi)}=\vert\tau(\chi)\vert^2=N$, one can rewrite this as $$\chi(n)=\frac{\chi(-1)\tau(\chi)}{N}\sum_{m\mod N}\overline{\chi(m)}\psi(nm)=\frac{\chi(-1)\tau(\chi)}{N}\sum_{m\mod N}\overline{\chi(m)}\psi(m)^n,$$ now we have that $$\sum_{n=1}^\infty \frac{\chi(n)}{n}=\frac{\chi(-1)\tau(\chi)}{N}\sum_{n=1}^{\infty}\sum_{m\mod N}\overline{\chi(m)}\frac{\psi(m)^n}{n}$$ $$=\frac{\chi(-1)\tau(\chi)}{N}\sum_{m\mod N}\overline{\chi(m)}\sum_{n=1}^{\infty}\frac{\psi(m)^n}{n}$$ Then use the above identity you can define $$L(1,\chi)=\sum_{n=1}^\infty\frac{\chi(n)}{n}=\frac{\chi(-1)\tau(\chi)}{N}\sum_{m\mod N}\overline{\chi(m)}\log(1-\psi(m))$$ Although the series on the left hand side is conditionally convergent, the right hand side gives you a finite number.
So in your question you are actually using $$\sum\frac{\chi(p)}{p}=\log L(1,\chi)+G(1)$$ to define the value of the series on the left hand side.