I have this in my notes where: $$\Phi(x) = \left\{ \begin{array}{ll} \Phi_I(x) = Ae^{k^{\prime}x} + Be^{-k^{\prime}x} &\text{if }x<0\\ \Phi_{II}(x) = C\sin(kx + \theta) &\text{if }0\le x\le a\\ \Phi_{III}(x) = De^{k^{\prime}x} + Ee^{-k^{\prime}x} &\text{if } x>0 \end{array} \right.$$ and it says that there is only a physically meaningful solution when $B=0$ and $D=0$.
Can someone explain why this is please? I'm just confused by it. I first think about complex numbers but there are none in the solution so what else could it be?
First, a typo in your question: I assume the wavefunction $\Phi_{III}(x)$ should be $De^{k^{\prime}x} + Ee^{-k^{\prime}x}$ in the region $x > a$, not $Ae^{k^{\prime}x} + Be^{-k^{\prime}x}$ in the region $x > 0$ as you've written.
States in quantum physics are only physically meaningful if they have a finite norm. In the case of a wavefunction in 1-D, this translates to the condition $$ \int_{-\infty}^\infty |\Phi(x)|^2 \, dx < \infty. $$ But if $D \neq 0$, the wavefunction will be approximately equal to $D e^{kx}$ as $x \to \infty$, and the integral will diverge as we take the bounds of integration to infinity. A similar thing will happen as $x \to -\infty$ if we have $B \neq 0$. Thus, we must have $B = D = 0$ for the wavefunction to have finite norm.