Let us consider a system of units in which mass and angular...

\left[M\right] = \left[Mass\right] = \left[{M}^{0}{L}^{0}{T}^{0}\right]
\left[J\right] = \left[Angularmomentum\right] = \left[M{L}^{2}{T}^{- 1}\right]
\left[L\right]$= \left[Length\right]$
[Now as angular momentum is dimensionless]
\left[M{L}^{2}{T}^{- 1}\right]$= \left[{M}^{0}{L}^{0}{T}^{0}\right]$
$∴ \left[{L}^{2}\right] = \left[T\right] <br >$∴ Power \left[P\right]$= \left[ML{T}^{- 2} ⋅ L{T}^{- 1}\right]$\left(asP = F ⋅ V\right)$<br >$= \left[M{L}^{2}{T}^{- 3}\right]
$= \left[{L}^{2}{L}^{- 6}\right]$
$⇒ [ P ] = [ {L}^{- 4} ] ⇒ ( A )$ is incorrect.
Energy/Work \left[W\right]$= \left[ML{T}^{- 2} ⋅ L\right]$ (using, $Work = Force \times displacement$
$= \left[{L}^{2}{L}^{- 4}\right] <br >$= \left[{L}^{- 2}\right] ⇒ \left(D\right) is correct.
Force $\left[F\right]$ = \left[ML{T}^{- 2}\right]$= \left[L ⋅ {L}^{- 4}\right]$ = \left[{L}^{- 3}\right]$⇒ ( B )$ is correct.
Linear momentum \left[P\right] = \left[ML{T}^{- 1}\right] = \left[L ⋅ {L}^{- 2}\right]
\left[p\right] = \left[{L}^{- 1}\right] ⇒ \left(C\right) is correct.