loc misbehaves when Period is at start of 3-level MultiIndex
#### Code Sample

```python
>>> pbc = pd.DataFrame({ 'p': [pd.Period('2017')], 'b':['b1'], 'c': ['c1'], 'v': [1]}).set_index(['p','b','c'])
>>> pbc.loc[(pd.Period('2017'), 'b1', 'c1'), 'v']
b   c
b1  c1    1
Name: v, dtype: int64
>>> pbc.loc[(pd.Period('2017'), 'wibble', 'c1'), 'v']
b   c
b1  c1    1
Name: v, dtype: int64
>>> pbc.loc[(pd.Period('2017'), 'b1', 'wobble'), 'v']
b   c
b1  c1    1
Name: v, dtype: int64
```

#### Problem description

I'd expect `loc` to return a scalar value when there's a unique match, rather than a `Series` with one item in it.

I'd also expect a `KeyError` when the second and third levels of the index don't match (the 'wibble' and 'wobble' selections). Instead it appears that the additional levels are ignored, and matches succeed or fail based on the period alone.

#### Expected Output

Everything works as I expect when I use strings instead of `Period`s:

```python
>>> ybc = pd.DataFrame({ 'y': ['2017'], 'b':['b1'], 'c': ['c1'], 'v': [1]}).set_index(['y','b','c'])
>>> ybc.loc[('2017', 'b1', 'c1'), 'v']
1
>>> ybc.loc[('2017', 'b1', 'wobble'), 'v']
...
KeyError: ('2017', 'b1', 'wobble')
```

Things also work as I'd expect when there are only two levels in the `MultiIndex` (or four, or five):

```python
>>> pb = pd.DataFrame({ 'p': [pd.Period('2017')], 'b':['b1'], 'v': [1]}).set_index(['p','b'])
>>> pb.loc[(pd.Period('2017'), 'b1'), 'v']
1
>>> pb.loc[(pd.Period('2017'), 'wibble'), 'v']
...
KeyError: (Period('2017', 'A-DEC'), 'wibble')
```

It also works if I put `Period` last in the three-level `MultiIndex`:

```python
>>> bcp = pd.DataFrame({ 'p': [pd.Period('2017')], 'b':['b1'], 'c': ['c1'], 'v': [1]}).set_index(['b','c', 'p'])
>>> bcp.loc[('b1', 'c1', pd.Period('2017')), 'v']
1
>>> bcp.loc[('b1', 'wibble', pd.Period('2017')), 'v']
...
KeyError: ('b1', 'wibble', Period('2017', 'A-DEC'))
```

... or second:

```
>>> bpc = pd.DataFrame({ 'p': [pd.Period('2017')], 'b':['b1'], 'c': ['c1'], 'v': [1]}).set_index(['b','p','c'])
>>> bpc.loc[('b1', pd.Period('2017'), 'c1'), 'v']
1
>>> bpc.loc[('b1', pd.Period('2017'), 'wibble'), 'v']
...
KeyError: ('b1', Period('2017', 'A-DEC'), 'wibble')
```

#### Output of ``pd.show_versions()``

<details>

[paste the output of ``pd.show_versions()`` here below this line]

INSTALLED VERSIONS
------------------
commit: None
python: 2.7.10.final.0
python-bits: 64
OS: Darwin
OS-release: 17.4.0
machine: x86_64
processor: i386
byteorder: little
LC_ALL: None
LANG: en_GB.UTF-8
LOCALE: None.None

pandas: 0.22.0
pytest: None
pip: 9.0.3
setuptools: 39.0.1
Cython: None
numpy: 1.14.2
scipy: None
pyarrow: None
xarray: None
IPython: None
sphinx: None
patsy: None
dateutil: 2.7.2
pytz: 2018.4
blosc: None
bottleneck: None
tables: None
numexpr: None
feather: None
matplotlib: None
openpyxl: None
xlrd: None
xlwt: None
xlsxwriter: None
lxml: None
bs4: None
html5lib: None
sqlalchemy: None
pymysql: None
psycopg2: None
jinja2: 2.10
s3fs: None
fastparquet: None
pandas_gbq: None
pandas_datareader: 0.5.0

</details>