The condition $|xy|\leq p$ requires that the selected substring $y$ to be pumped is not too far from the start of the string.
Consider the language $L = {x |x \in {0,1}^* x\ has\ equal\ number\ of\ 0's\ and\ 1's }$ which can be shown to be a non-regular language via the pumping lemma. To show that this is not regular the usual example string used is $0^p1^p$. With the 2nd condition, no matter how we divide this string, we only get 0's for $y$, so the lemma fails for this. If we just require $1 \leq|y|\leq p$ then we are free to select anywhere in the substring, say $y = 0^{p/2}1^{p/2}$ and pump the substring and pass the lemma.
The requirement of selecting $y$ not to be far from the start of the string has something to do with the proof that the pumping lemma exists in the first place. So you might want to check it out from some book like Sipser or whatever you are using or here.
Edit: I decided to give my own explanation in case you do not want to read the references
In the proof of the pumping lemma, given some regular language $L$, $p$ is the number of states of the smallest DFA that recognizes $L$. Say that there exists $w\in L$ such that $|w| \geq p$. Remember that before we even begin consuming $w$, we are already at the start state so technically we have already entered 1 state and there are $p-1$ remaining states that we have not entered yet. But $w$ contains $\geq p$ symbols and reading a symbol will result to entering a state, so it must be that at least one state was entered more that once. This is where the pigeonhole principle is applied.
From above, it must be that the sequence of states entered to consume $w$, from the initial to the final state, contains a cycle. 399
Now, string $w=xyz$ because $x$ represents the substring consumed before entering the cycle, $y$ is the substring consumed in the cycle, and $z$ is the consumed after the cycle.
Given a language $L$ and a string $w$ from it, you do not necessarily need to find the smallest DFA to get $p$. It is known to be there, assuming that $L$ is regular. What you are trying to do is to guess a correct partition of $w$. Finding the correct partition is equivalent to finding the state $q_i$ where the cycle will begin. Of course, when you find $q_i$ you cannot start the DFA there since you must begin at the start state (unless of course $q_i$ is the start state which implies that $|x|=0$). So your simplified bound of $1 \leq y \leq p$ is not enough. Essentially, the substring $x$ is like your offset from the start state to $q_i$. Since there are at most $p$ states, and you are initially at the start state, you can try to visit at most $p-1$ states until you find $q_i$ which implies that the length of $x$ is at most $p-1$. Since $|y| \geq 1$, the length of $xy \leq p$ 399