Error probability for orthogonal signaling with unequal symbol probabilities
I haven't checked the details fully but the OP's revised solution is correct. Once upon a time, I had posted a to a class webpage from 1998 where I had some lecture notes giving the answer 
but the link is now defunct because my University seems to have deleted that webpage from its servers.  Anyway, the answer that I came with to the OP's question is
$$P_e = \pi_0 Q\left(\mathsf{SNR} -(2\cdot\mathsf{SNR})^{-1}\ln\left(\frac{\pi_1}{\pi_0}\right)\right) + \pi_1 Q\left(\mathsf{SNR} +(2\cdot\mathsf{SNR})^{-1}\ln\left(\frac{\pi_1}{\pi_0}\right)\right) \tag{1}$$
where $\mathsf{SNR} = \dfrac{\Vert s_0-s_1\Vert}{\sqrt{2N_0}} = \sqrt{\dfrac{E_b}{N_0}}$ in this case. Note that when $\pi_1 &gt; \pi_0$, $\ln\left(\frac{\pi_1}{\pi_0}\right) &gt; 0$, and so $$Q\left(\mathsf{SNR} +(2\cdot\mathsf{SNR})^{-1}\ln\left(\frac{\pi_1}{\pi_0}\right)\right) &lt; Q\left(\mathsf{SNR} -(2\cdot\mathsf{SNR})^{-1}\ln\left(\frac{\pi_1}{\pi_0}\right)\right),$$
that is, in the weighted sum of two error probabilities in $(1)$, the larger weight $\pi_1$ weights the smaller error probability, which is exactly how it should be.