All of the answers here, including the accepted one, are conspicuously confused. I down-voted the accepted answer but downvotes of users who lack reputation in this "community" are not counted. I have a reputation of more than 200,000 (two-hundred-thousand) on math.stackexchange.com and I have a Ph.D. in statistics, but none of that counts here.
The question says $P(B\mid A)$ is the likelihood, but it ought to say instead that $A\mapsto P(B\mid A)$ is the likelihood. I.e. it's this probability as a function of $A,$ not of $B.$
It also says $P(B\mid A)P(A)$ is the conditional probability. That is not right. $P(B\mid A)$ is a conditional probability. $P(B\mid A)P(A)$ is $P(A\ \&\ B).$
Now suppose you have
$$
\Pr(A_1) = \frac 1 {20}, \quad \Pr(A_2) = \frac 2 {20}, \quad \Pr(A_3) = \frac{17}{20}
$$
and $A_1,A_2,A_3$ are mutually exclusive. This is a prior probability distribution.
Further suppose that
$$
\Pr(B\mid A_1) = \frac 9 {10}, \quad \Pr(B\mid A_2)= \frac 2 3, \quad \Pr(B\mid A_3) = \frac 1 2.
$$
Note well:
These three probabilities do not add up to $1.$ What is expressed here is not a probability distribution.
It is a function of $\text{“}A\text{''},$ a variable whose value may be either $A_1,$ $A_2,$ or $A_3.$
The likelihood is $\Pr(B\mid A)$ as a function of $A,$ not of $B.$
Bayes's theorem says: When one multiplies (pointwise) the prior probability distribution by the the likelihood and then normalizes, one gets the posterior probability distribution, i.e. the distribution conditional on the data. The "data" is the observed event $B.$
Thus
\begin{align}
& \left( \tfrac 1 {20}, \tfrac 2 {20}, \tfrac{17}{20} \right) \times\left( \tfrac 9{10}, \tfrac 2 3, \tfrac 1 2 \right) \\[8pt]
= {} & \left( \tfrac 9 {200}, \tfrac 2 {60}, \tfrac{17}{40} \right) \\[8pt]
\propto {} & \left( 27, 20, 255 \right) \\
& \text{(Here I multiplied all three components} \\
& \phantom{(} \text{by 600, which is the l.c.m. of the denominators.)} \\[8pt]
\propto {} & \left( \tfrac{27}{302}, \tfrac{20}{302}, \tfrac{255}{302} \right) \\
& \text{(Here I normalized, i.e. divided by} \\
& \phantom{(} 27+20+255 = 302 \text{ so that the} \\
& \phantom{(} \text{sum of the three components is $1.$)} \\[8pt]
= {} & \left( \Pr(A_1\mid B), \Pr(A_2\mid B), \Pr(A_3\mid B) \right) \\[8pt]
= {} & \textbf{the posterior probability distribution.}
\end{align}
Often one sees something like
$$
X\sim\operatorname{Binomial}(4,p),
$$
so that, for example
$$
\Pr(X=2) = \binom 4 2 p^2 (1-p)^{4-2}.
$$
Then if the data consists of the observation that $X=2,$ then the likelihood function is
$$
L(p) = \binom 4 2 p^2(1-p)^{4-2}.
$$
This is a function of $p,$ not of a variable whose values can be the possible values of $X,$ which in this case are $0,1,2,3,4.$