Can't convert from PyPtr to related type

Hello everyone and thanks for the great library. I was trying to use rdflib of Python in PyCall.jl. I'm using macos, Julia 1.9.1 and using `ENV["Python"]=""` with Conda.jl. With this code,

```julia
using PyCall

# Create an empty graph
graph = pyimport("rdflib").Graph()

# Define the namespace
ex = "http://example.org/"

# Define the triples
subject = pyimport("rdflib").URIRef(ex * "subject")
predicate = pyimport("rdflib").URIRef(ex * "predicate")
object = pyimport("rdflib").Literal("object")

# Add the triples to the graph
graph.add((subject, predicate, object))
```

I got the following output
```julia
ERROR: PyError ($(Expr(:escape, :(ccall(#= /Users/hiiroo/.julia/packages/PyCall/ilqDX/src/pyfncall.jl:43 =# @pysym(:PyObject_Call), PyPtr, (PyPtr, PyPtr, PyPtr), o, pyargsptr, kw)))))
AssertionError('Subject http://example.org/subject must be an rdflib term')
File "/Users/hiiroo/.julia/conda/3/lib/python3.7/site-packages/rdflib/graph.py", line 532, in add
assert isinstance(s, Node), "Subject %s must be an rdflib term" % (s,)
```

Reproduce the macOS Julia 1.9.1 example with PyCall.jl, Conda.jl, and rdflib. Start at PyCall's src/pyfncall.jl:43 and the rdflib assertion in graph.py, then trace how the URIRef and Literal arguments are converted. Done means the graph.add call accepts the created rdflib terms without the assertion error.

Scritto dal modello di indicizzazione a partire dal testo della issue.