It seems to me that you can show that no separable Banach space $X$ is P-complete as follows. Pick any bounded separated sequence $\{x_n\}$ in $X$ and pick a dense sequence $\{y_i\}$ in $X$. Pick a subsequence in $\{x_n\}$ for which $\|x_n-y_1\|$ converges. Then from this subsequence pick further subsequence for which $\|x_n-y_2\|$ converges. So on. After doing this for all $i$, pick a diagonal subsequence $\{x_{n(k)}\}_{k=1}^\infty$ and show that it satisfies the desired conditions.

It seems to me that you can show that no infinite-dimensional separable Banach space $X$ is P-complete as follows. Pick any bounded separated sequence $\{x_n\}_{n=1}^\infty$ in $X$ and pick a dense sequence $\{y_i\}$ in $X$. Pick a subsequence in $\{x_n\}$ for which $\|x_n-y_1\|$ converges. Then from this subsequence pick further subsequence for which $\|x_n-y_2\|$ converges. So on. After doing this for all $i$, pick a diagonal subsequence $\{x_{n(k)}\}_{k=1}^\infty$ and show that it satisfies the desired conditions.