 Solution

For the amplitude continue to $k=i\kappa$ on the decaying physical sheet. Then $\zeta =q$ and $D=\cos (qa)+({\kappa }^2-q^2)\sin (qa)/(2\kappa q)$. The double-angle identity factors its zero condition into the even and odd matching conditions $\kappa =q\tan (qb)$ and $\kappa =-q\cot (qb)$. Thus physical upper-half-plane amplitude poles are exactly the bound energies.

The literal statement for the probability requires correction. The displayed real-axis expression continued algebraically is $1/[D(k)D(-k)]$, and includes poles of the opposite-boundary-condition amplitude too. A concrete counterexample uses ${\hslash }^2/(2m)=1$, $a=2$, $q=1$, $\kappa =\cot 1$ and $U_0=1+{\cot }^{2}1$. Its continued probability denominator vanishes at $k=i\cot 1$, because $D(-k)=0$. But neither $\kappa =\tan 1$ nor $\kappa =-\cot 1$ holds, so $E=-{\cot }^{2}1$ is not a bound energy. Keeping the amplitude and the decaying boundary condition removes this spurious identification.