\(P=\left[M L^{-1} T^{-2}\right]\)

\(a\) will have same unit as \(t^{2} .\) So, \(a=\left[T^{2}\right]\)

\(b=\left[\frac{a-t^{2}}{p x}\right]=\frac{\left[T^{2}\right]}{\left[M L^{-1} T^{-2} L\right]}=\left[M^{-1} L^{0} T^{4}\right]\)

\(\frac{a}{b}=\frac{\left[T^{2}\right]}{\left[M^{-1} L^{0} T^{4}\right]}=\left[M L^{0} T^{-2}\right]\)