`pip.parse(python_version=)` should not be mandatory

# 🐞 bug report

### Affected Rule

Bzlmod `pip.parse()`

### Is this a regression?

Yes, was not a problem in WORKSPACE.

### Description

The arg documentation says the following:
https://github.com/bazelbuild/rules_python/blob/711186f144af06b431bd416b2d742874de3a2dea/python/private/bzlmod/pip.bzl#L391-L398

It specifically describes `If not specified, then the default Python version (as set by the root module or rules_python) will be used`. So this attribute should be optional but currently is mandatory.

## 🔬 Minimal Reproduction

`pip.parse()` without `python_version`.

## 🔥 Exception or Error

```
ERROR: in tag at /MODULE.bazel:42:10, mandatory attribute python_version isn't being specified
```

## 🌍 Your Environment

**Operating System:**

  

Linux

  

**Output of `bazel version`:**

  

2024/01/19 15:28:42 Warning: used fallback version "6.3.2"

Bazelisk version: v1.19.0

INFO: Running bazel wrapper (see //tools/bazel for details), bazel version 6.1.2 will be used instead of system-wide bazel installation.

Build label: 6.1.2

Build target: bazel-out/k8-opt/bin/src/main/java/com/google/devtools/build/lib/bazel/BazelServer_deploy.jar

Build time: Tue Apr 18 15:29:54 2023 (1681831794)

Build timestamp: 1681831794

Build timestamp as int: 1681831794

  

**Rules_python version:**

  

0.27.1

  

**Anything else relevant?**

Lisez python/private/bzlmod/pip.bzl autour de l’attribut python_version documenté aux lignes 391-398. Reproduisez l’appel minimal de MODULE.bazel utilisant pip.parse() sans python_version, puis vérifiez que l’appel réussit et utilise la version de Python par défaut documentée.

Rédigé par le modèle d'indexation à partir du texte de l'issue.