Particle-hole symmetry and the sign of the superconducting gap Time-reversal (TR) symmetry leads to topological insulator property. As expected, the topological invariant depends on TR (Pfaffian) operator. 
TR+particle-hole symmetry leads to topological superconductor. The topological invariant depends on the Chern number and the sign of the gap. 
Is it physically correct to say that the Chern number represents the TR symmetry of the band and the sign of the gap represents the particle-hole symmetry ? If yes, what is the physical intuition behind the representation of the particle-hole symmetry with sign of the gap ? 
 A: No, the sign of the gap does not represent particle-hole symmetry. The superconducting gap simply being non-zero automatically encodes the existence of particle-hole symmetry. Variations in the sign of the gap in the Brillouin zone (BZ) determines whether a superconductor is topologically trivial or non-trivial. However, topological or not, a superconductor will possess particle-hole symmetry.
I believe some background information might provide some context to the general readers. For a time-reversal invariant 3D topological superconductor the topological invariant or winding number can be written as $$N_{W}=\frac{1}{2}\sum_{s}{\rm sgn}(\delta_{s})C_{s}$$ where $$C_{s}=\frac{1}{2\pi}\int_{{\rm FS}_{s}}d\Omega^{ij}\left[\partial_{i}a_{sj}({\bf k})-\partial_{j}a_{si}({\bf k})\right]$$ is a Chern-number-like quantity evaluated over the $s^{{\rm th}}$ Fermi surface (which is 2D for a 3D superconductor), $a_{si}=-{\rm i}\left\langle s{\bf k}\right|\partial/\partial k_{i}\left|s{\bf k}\right\rangle$, where $\left|s{\bf k}\right\rangle$ is a Bloch state on the $s^{\rm th}$ Fermi surface, ${\rm sgn}(\delta_{s})$ represents the sign of the gap on the $s^{{\rm th}}$ Fermi surface, and $i,j=x,y,z$. The $s$ different Fermi surfaces also need to be disconnected in the BZ. Corresponding expressions can be found in lower dimensions by a formal procedure called “dimensional reduction.” However, let’s focus on the 3D case for now.
If our system obeys time-reversal symmetry then $$\sum_{s}C_{s}=0$$ For a conventional $s$-wave superconductor, the sign, by definition, is constant throughout the BZ. This always implies $N_{W}=0$ in a time-reversal invariant system. For example, for a system with 2 disconnected Fermi surfaces, (say) we have $C_{\pm}=\pm 1$. For an $s$-wave superconductor ${\rm sgn}(\delta_{\pm})=+1$ (or $-1$ for both). Then we obviously have $$N_{W}=\frac{1}{2}(+1)(+1)+\frac{1}{2}(+1)(-1)=0$$ But if ${\rm sgn}(\delta_{\pm})=\pm 1$ (say fully gapped $p$-wave), then $$N_{W}=\frac{1}{2}(+1)(+1)+\frac{1}{2}(-1)(-1)=1$$ In summary, both cases have particle-hole symmetry. The sign of the gap simply indicates topological character.
Further details regarding the above formalism can be found in

Xiao-Liang  Qi, Taylor L. Hughes, and Shou-Cheng Zhang. “Topological invariants for the Fermi surface of a time-reversal-invariant superconductor.” Physical Review B 81, no. 13 (2010): 134508. (arXiv)

