If momentum (P), area (A) and time (T) are taken to be the fundamental quantities then the dimensional formula for energy is:

We know that in the usual MLT system the dimensional formula for mechanical energy (work) is $$[E]=[M\,L^{2}\,T^{-2}]$$, because work is force times distance and force itself is mass times acceleration.

Now the problem tells us to treat momentum $$P$$, area $$A$$ and time $$T$$ as the new fundamental quantities. Therefore we must rewrite the above $$[M\,L^{2}\,T^{-2}]$$ in terms of $$P$$, $$A$$ and $$T$$ only.

First we recall the standard dimensional expressions of the given fundamental quantities in the MLT system:

Momentum is defined as mass times velocity, so $$[P]=[M\,L\,T^{-1}].$$

Area is the product of two lengths, hence $$[A]=[L^{2}].$$

Time is already fundamental, so $$[T]=[T].$$

Let us now assume that the energy dimension can be built as a product of powers of these three fundamental quantities. We therefore write $$[E]=P^{x}\,A^{y}\,T^{z},$$ where $$x$$, $$y$$ and $$z$$ are the unknown exponents to be determined.

Next, we substitute the MLT expressions of $$P$$, $$A$$ and $$T$$ into this assumed form:

$$$ [E]=\bigl(M^{1}L^{1}T^{-1}\bigr)^{x}\; \bigl(L^{2}\bigr)^{y}\; \bigl(T^{1}\bigr)^{z}. $$$

Collecting the powers of each fundamental symbol $$M$$, $$L$$ and $$T$$ separately, we get

$$$ [E]=M^{\,x}\; L^{\,x+2y}\; T^{\,-x+z}. $$$

This combined exponent form must equal the original energy dimension $$M^{1}L^{2}T^{-2}$$, so we equate the corresponding powers term by term:

For mass $$M$$: $$$ x = 1. $$$

For length $$L$$: $$$ x + 2y = 2. $$$

For time $$T$$: $$$ -x + z = -2. $$$

We now solve these three simple linear equations step by step.

From the first equation we immediately have $$$ x = 1. $$$

Substituting $$x=1$$ into the second equation gives $$$ 1 + 2y = 2 \quad\Longrightarrow\quad 2y = 1 \quad\Longrightarrow\quad y = \frac{1}{2}. $$$

Substituting $$x=1$$ into the third equation gives $$$ -1 + z = -2 \quad\Longrightarrow\quad z = -1. $$$

Thus the required exponents are $$$ x = 1,\qquad y = \tfrac12,\qquad z = -1. $$$

Putting these values back into the assumed form $$P^{x}A^{y}T^{z}$$, we obtain

$$$ [E]=P^{1}\,A^{1/2}\,T^{-1}. $$$

Therefore the dimensional formula for energy, when momentum, area and time are chosen as fundamental quantities, is written as $$[PA^{1/2}T^{-1}].$$

Looking at the given options, this matches Option C.

Hence, the correct answer is Option C.