Solution:

a) The answer is "Yes". For $P(x)$ to be balanced, $x+a$ and $x+b$ must either both have an even or both have an odd number of prime factors. With respect to this property, however, there are only $2^{50}$, hence finitely many, different patterns among 50 consecutive natural numbers. Therefore there exist two natural numbers $m$ and $n$ that are starting numbers of two identical such patterns. With $a = m-1$ and $b = n-1$, the balancedness of $P(1), P(2), \ldots, P(50)$ follows.

b) We assume that under the given hypothesis there exist two distinct natural numbers $a$ and $b$; without loss of generality let $b > a$. Then for every natural number $m > a$ we have that $P(m-a) = m(m + b - a)$ is balanced. The evenness or oddness of the number of prime factors thus occurs periodically for natural numbers greater than $a$, with period length $b-a$. In particular, all multiples of $b-a$ are also of the same type, as soon as they are greater than $a$. However, such a multiple $k(b-a)$ has one fewer prime factor than $2k(b-a)$, a contradiction to the assumption. Hence $a = b$.