DPSK vs PSK Error Probability
The flaw in the problem posed starts right from the block diagram  that you have posted: it is not the case in DPSK receiver that the RF signal is multiplied with a delayed version of itself and then passed through a low-pass filter followed by slicing etc. 

Instead, the RF signal is "demodulated" through a variant of a standard coherent QPSK receiver whose local oscillator is assumed to be synchronized in frequency to the incoming carrier signal but not necessarily synchronized in phase. Thus, the incoming $$r(t) = g(t)\cos(2\pi f_c t+\theta) + n(t), 0 \leq t &lt; T,$$ where $g(t)$ is the baseband pulse, $T$ the bit duration, $\theta$ the unknown RF phase, and $n(t)$ the AWGN (or band limited AWGN if you like to think that the incoming signal has been bandpass-filtered, e.g. as in a tuned amplifier in an IF stage of a superheterodyne system), is multiplied by $2\cos(2\pi f_c t)$ in the I branch and separately by $-2\sin(2\pi f_c t)$ in the Q branch. Since 
\begin{align}
2\cos A \cos B &amp;= \cos(A+B) + \cos(A-B)\\
2\sin A \cos B &amp;= \sin(A+B) + \sin(A-B)
\end{align}
we get that the signal parts of the two mixer outputs are
\begin{align}
2\cos(2\pi f_c t) g(t)\cos(2\pi f_c t+\theta) &amp;= g(t)\cos(4\pi f_c t+\theta) + g(t)\cos(-\theta)\\
-2\sin(2\pi f_c t) g(t) \cos(2\pi f_c t+\theta)&amp;= -g(t)\sin(4\pi f_c t+\theta) - g(t)\sin(-\theta).
\end{align}
The double-frequency terms in the mixer outputs can be eliminated explicitly by low-pass filtering the results before doing anything else, or they get eliminated implicitly when we do matched filtering on the two mixer outputs. Now, matched filtering is typically done by via correlation
(multiply by $g(t)$ and integrate from $0$ to $T$) and thus we effectively get a complex-valued decision variable $Xe^{j\theta} + \mathcal CN(0,\sigma^2)$ at the end of the integration period. We can't make a decision just on this decision variable because we don't know the value of $\theta$; depending on what $\theta$ is, the decision might be wholly bass ackwards from what it should be! But what we can do is save the decision variable and use it $T$ seconds later to answer the question


  Did the incoming signal change phase from the unknown $\theta$ during $[0,T)$ to (also unknown) $\theta+\pi$ during $[T,2T)$, or did the phase stay the same unknown $\theta$ during $[T,2T)$? 


The final expression for the error probability of DBPSK is quite simple:
$$P_{\text{e, DBPSK}} = \frac 12 e^{-E_b/N_0}$$
but, as you say, is harder to derive than the $Q
\left(\sqrt{\frac{2E_b}{N_0}}\right)$ error probability for coherent BPSK.

Turning to the question of why the trick of using the coherent BPSK error formula as a guide and simply replacing the SNR for BPSK with the SNR for DBPSK in the error probability formula doesn't give the right answer for the DBPSK error probability, the issue is that the two systems make decisions differently and there is no obvious reason why the error probabilities of the two systems should be given by the same formula in terms of SNR. Note, for example, that $P_e = Q(\sqrt{2E_b/N_0})$ for coherent BPSK but only $Q(\sqrt{E_b/N_0})$ for coherent FSK.

Finally, note that for $x &gt; 0$,
$Q(x)  &lt; \frac 12 \exp\left(-\frac{x^2}{2}\right)$
and so, setting $x=\sqrt{2E_b/N_0}$, we get that
$$P_{\text{e, BPSK}} = Q(\sqrt{2E_b/N_0}) &lt; P_{\text{e, DBPSK}} = \frac 12 e^{-E_b/N_0}$$