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. However, there are only $2^{50}$, i.e., a finite number of different patterns among 50 consecutive natural numbers. Therefore, there exist two natural numbers $m$ and $n$ such that they are the starting points of two identical such patterns. With $a=m-1$ and $b=n-1$, the balance of $P(1), P(2), \ldots, P(50)$ follows.
b) We assume that there are two different natural numbers $a$ and $b$ under the given condition; without loss of generality, let $b>a$. Then for every natural number $m>a$, $P(m-a)=m(m+b-a)$ is balanced. The evenness or oddness of the number of prime factors thus occurs periodically with period $b-a$ for natural numbers greater than $a$. In particular, all multiples of $b-a$ are of the same type, provided they are greater than $a$. However, such a multiple $k(b-a)$ has one fewer prime factor than $2k(b-a)$, leading to a contradiction with the assumption. Therefore, $a=b$.
Hint: The contradiction can be derived in various ways. Often, theorems about prime numbers in arithmetic sequences or suitable square numbers were used. For part a), 4 points were awarded, and for part b), 6 points were awarded.