According to the principle of dimensional homogeneity, terms added or subtracted must have the same dimensions. In the first parenthesis, $P$ is added to $\frac{a}{V^2}$. Therefore, the dimension of $\frac{a}{V^2}$ must be the same as the dimension of pressure ($P$). Dimensions of pressure $P = \frac{\text{Force}}{\text{Area}} = \frac{[MLT^{-2}]}{[L^2]} = [ML^{-1}T^{-2}]$. Dimensions of volume $V = [L^3]$, so $V^2$ has dimensions $[L^6]$. Equating the dimensions: $[\frac{a}{V^2}] = [P]$ $[a] = [P][V^2]$ $[a] = [ML^{-1}T^{-2}][L^6] = [ML^{(-1+6)}T^{-2}] = [ML^5T^{-2}]$.