Choose x = 0P+11p, which is obviously in L.
Then x = u v w, |v| ≥1, |uv| ≤p, and every u vmw, for any m ≥0, is in L.
Considering m = 0, we know that u w is in L.
v consists of just 0s, and contains at least one 0.
So removing v removes at least one 0, which yields a string that is not in L