What is the physical significance of the Dual Space in QM?

Kets do not represent wave functions, so this may contribute to your idea of there being an imbalance between bras and kets. Instead, we do not get a wave function associated with a state (ket, $\left|\alpha\right>$) until we project it onto a basis, such as the position- or momentum-space bases, revealing that the wave function is actually a scalar:

$\psi(x) = \left<x|\alpha\right>$, $\phi(p) = \left<p|\alpha\right>$.

This can be likened to the scenario of a vector, $\vec{v}$, which has an existence and meaning independent of any basis, as can be seen geometrically with an arrow. It is only when we choose a particular basis, such as $\hat{i}, \hat{j}, \hat{k}$, that we can specify the vector by its (scalar) components. Similarly, the wave functions (in $x$, $p$, or some other basis) merely specify the components of a given ket (for a nice discussion related to this, see the opening chapter of Shankar's Principles of Quantum Mechanics). There is nothing stopping us from projecting a bra onto a basis:

$\psi^*(x) = \left<\alpha|x\right> = \left<x|\alpha\right>^*,\ \phi^*(p) = \left<\alpha|p\right> = \left<p|\alpha\right>^*$.

What this gives us, though, is the complex conjugate of our previous wave functions. The physical meaning of why the complex conjugate shows up with the bra here is related to our conception of probability being a non-negative measure: since quantum mechanics necessitates the use of complex numbers, in order to guarantee that $\left<\alpha|\alpha\right> \ge 0$, the dual to a ket has to involve complex conjugation ($z^* z \ge 0$ in general). As previous comments have indicated, the only remaining difference is that between column vectors and row vectors, which is arguably a convention.