According to the principle of dimensional homogeneity, each term in the given equation must have the same dimensions.

Given equation: $(P+\frac{A^{2}}{B})+\frac{1}{2}\rho V^{2}=$ constant

Therefore, the dimensions of $P$, $\frac{A^{2}}{B}$, and $\frac{1}{2}\rho V^{2}$ must be the same.

1. Dimensions of Pressure ($P$):

$[P] = ML^{-1}T^{-2}$

2. Dimensions of Density ($\rho$) and Velocity ($V$):

$[\rho] = ML^{-3}$

$[V] = LT^{-1}$

Therefore, the dimensions of $\frac{1}{2}\rho V^{2}$ are:

$[\frac{1}{2}\rho V^{2}] = [ML^{-3}][LT^{-1}]^{2} = ML^{-3}L^{2}T^{-2} = ML^{-1}T^{-2}$

3. Since $[P] = [\frac{A^{2}}{B}]$, we have:

$[\frac{A^{2}}{B}] = ML^{-1}T^{-2}$

We need to find the dimensions of $\frac{A}{B}$. Let's denote it as $[X] = [\frac{A}{B}]$

Then, $[\frac{A^{2}}{B}] = [A] \cdot [\frac{A}{B}] = [A] \cdot [X] = ML^{-1}T^{-2}$

4. Also, since $[P] = [\frac{1}{2}\rho V^{2}]$, we can equate the dimensions:

$P = \frac{A^2}{B}$

$\frac{A^2}{B} = \rho V^2$

$\frac{A^2}{B} = ML^{-1}T^{-2}$

Now, let's consider $\frac{A}{B} = X$. Then $A = BX$. Substituting this into the previous equation:

$\frac{(BX)^2}{B} = ML^{-1}T^{-2}$

$B X^2 = ML^{-1}T^{-2}$

We also know that $\frac{A^2}{B}$ has the same dimensions as pressure, so $\frac{A^2}{B} = ML^{-1}T^{-2}$.

Let's find the dimensions of A first. Since $\frac{A^2}{B}$ has the same dimensions as pressure and $\frac{1}{2}\rho V^2$, we have:

$\frac{A^2}{B} \sim ML^{-1}T^{-2}$

Now, consider the term $\frac{1}{2}\rho V^2$. Its dimensions are $ML^{-1}T^{-2}$. Therefore, $\frac{A^2}{B}$ must also have these dimensions.

Let's assume $\frac{A}{B} = X$. Then $A = BX$. So, $\frac{(BX)^2}{B} = BX^2 = ML^{-1}T^{-2}$.

We need to find X. Let's try another approach.

Since $P$ and $\frac{1}{2}\rho V^2$ have the same dimensions, we can write:

$P = \frac{A^2}{B}$

$ML^{-1}T^{-2} = \frac{A^2}{B}$

We want to find $\frac{A}{B}$. Let's rewrite the equation as:

$A^2 = B(ML^{-1}T^{-2})$

$A = \sqrt{B(ML^{-1}T^{-2})}$

Now, $\frac{A}{B} = \frac{\sqrt{B(ML^{-1}T^{-2})}}{B} = \sqrt{\frac{ML^{-1}T^{-2}}{B}}$

This approach doesn't seem to simplify things. Let's go back to $BX^2 = ML^{-1}T^{-2}$.

Since $\frac{A^2}{B}$ has the same dimensions as $\rho V^2$, we have $\frac{A^2}{B} = ML^{-1}T^{-2}$.

Let $X = \frac{A}{B}$. Then $A = BX$. Substituting this into the equation above:

$\frac{(BX)^2}{B} = ML^{-1}T^{-2}$

$BX^2 = ML^{-1}T^{-2}$

We need to find $X = \frac{A}{B}$. Let's try to express $A$ in terms of known quantities.

From $\frac{A^2}{B} = ML^{-1}T^{-2}$, we have $A^2 = B(ML^{-1}T^{-2})$.

Now, let's consider the dimensions of $B$. We know that $\frac{A^2}{B}$ has the dimensions of pressure, which is $ML^{-1}T^{-2}$.

Let's assume $A$ has dimensions $M^aL^bT^c$ and $B$ has dimensions $M^xL^yT^z$.

Then $\frac{(M^aL^bT^c)^2}{M^xL^yT^z} = ML^{-1}T^{-2}$.

$\frac{M^{2a}L^{2b}T^{2c}}{M^xL^yT^z} = ML^{-1}T^{-2}$.

$M^{2a-x}L^{2b-y}T^{2c-z} = ML^{-1}T^{-2}$.

So, $2a-x = 1$, $2b-y = -1$, and $2c-z = -2$.

We want to find $\frac{A}{B}$, which has dimensions $\frac{M^aL^bT^c}{M^xL^yT^z} = M^{a-x}L^{b-y}T^{c-z}$.

Let $a-x = p$, $b-y = q$, and $c-z = r$. We want to find $M^pL^qT^r$.

We have $2a-x = 1$, so $a = \frac{1+x}{2}$. Then $p = a-x = \frac{1+x}{2} - x = \frac{1-x}{2}$.

We have $2b-y = -1$, so $b = \frac{-1+y}{2}$. Then $q = b-y = \frac{-1+y}{2} - y = \frac{-1-y}{2}$.

We have $2c-z = -2$, so $c = \frac{-2+z}{2}$. Then $r = c-z = \frac{-2+z}{2} - z = \frac{-2-z}{2}$.

So, we want to find $M^{\frac{1-x}{2}}L^{\frac{-1-y}{2}}T^{\frac{-2-z}{2}}$.

Let's consider the given options. We have $ML^{-1}T^{-4}$. This means $x = -1, y = 1, z = 8$.

Then $\frac{1-x}{2} = \frac{1-(-1)}{2} = 1$. $\frac{-1-y}{2} = \frac{-1-1}{2} = -1$. $\frac{-2-z}{2} = \frac{-2-8}{2} = -5$.

So, we have $ML^{-1}T^{-5}$, which is not in the options.

Let's try to find the dimensions of A and B separately. We know that $\frac{A^2}{B}$ has the dimensions of pressure, $ML^{-1}T^{-2}$.

Let's assume $B$ has the dimensions of area, $L^2$. Then $A^2$ has the dimensions of $ML^{-1}T^{-2}L^2 = MLT^{-2}$. So, $A$ has the dimensions of $M^{1/2}L^{1/2}T^{-1}$.

Then $\frac{A}{B}$ has the dimensions of $\frac{M^{1/2}L^{1/2}T^{-1}}{L^2} = M^{1/2}L^{-3/2}T^{-1}$. This is not in the options.

Since $P$ and $\frac{1}{2}\rho V^2$ have the same dimensions, we have $ML^{-1}T^{-2}$. Thus, $\frac{A^2}{B}$ must also have these dimensions.

Let's assume $A$ has dimensions $M^aL^bT^c$ and $B$ has dimensions $M^xL^yT^z$. Then $\frac{A}{B}$ has dimensions $M^{a-x}L^{b-y}T^{c-z}$.

We know that $\frac{A^2}{B}$ has dimensions $ML^{-1}T^{-2}$. So, $\frac{(M^aL^bT^c)^2}{M^xL^yT^z} = ML^{-1}T^{-2}$.

This means $2a-x = 1$, $2b-y = -1$, and $2c-z = -2$.

We want to find $\frac{A}{B}$, which has dimensions $M^{a-x}L^{b-y}T^{c-z}$.

Let's try option (A): $MLT^{-4}$. Then $a-x = 1$, $b-y = 1$, and $c-z = -4$.

We have $2a-x = 1$, so $a = \frac{1+x}{2}$. Then $a-x = \frac{1+x}{2} - x = \frac{1-x}{2} = 1$. So, $1-x = 2$, which means $x = -1$.

We have $2b-y = -1$, so $b = \frac{-1+y}{2}$. Then $b-y = \frac{-1+y}{2} - y = \frac{-1-y}{2} = 1$. So, $-1-y = 2$, which means $y = -3$.

We have $2c-z = -2$, so $c = \frac{-2+z}{2}$. Then $c-z = \frac{-2+z}{2} - z = \frac{-2-z}{2} = -4$. So, $-2-z = -8$, which means $z = 6$.

So, $B$ has dimensions $M^{-1}L^{-3}T^6$. Then $A$ has dimensions $M^0L^{-2}T^2$.

Then $\frac{A^2}{B}$ has dimensions $\frac{(M^0L^{-2}T^2)^2}{M^{-1}L^{-3}T^6} = \frac{L^{-4}T^4}{M^{-1}L^{-3}T^6} = ML^{-1}T^{-2}$. This is correct.

So, $\frac{A}{B}$ has dimensions $\frac{M^0L^{-2}T^2}{M^{-1}L^{-3}T^6} = MLT^{-4}$.