interface IInterface
{
}

class TheClass implements IInterface
{
}

$cls = new TheClass();
if ($cls instanceof IInterface) {
    echo "yes";
}

You can use the "instanceof" operator. To use it, the left operand is a class instance and the right operand is an interface. It returns true if the object implements a particular interface.

nlaq points out that instanceof can be used to test if the object is an instance of a class that implements an interface.

But instanceof doesn't distinguish between a class type and an interface. You don't know if the object is a class that happens to be called IInterface.

You can also use the reflection API in PHP to test this more specifically:

$class = new ReflectionClass('TheClass');
if ($class->implementsInterface('IInterface'))
{
  print "Yep!\n";
}

See http://php.net/manual/en/book.reflection.php

Update

The is_a function is missing here as alternative.

I did some performance tests to check which of the stated ways is the most performant.

Results over 100k iterations

      instanceof [object] took   7.67 ms | +  0% | ..........
            is_a [object] took  12.30 ms | + 60% | ................
             is_a [class] took  17.43 ms | +127% | ......................
class_implements [object] took  28.37 ms | +270% | ....................................
       reflection [class] took  34.17 ms | +346% | ............................................

Added some dots to actually "feel" see the difference.

Generated by this: https://3v4l.org/8Cog7

Conclusion

In case you have an object to check, use instanceof like mentioned in the accepted answer.

In case you have a class to check, use is_a.

Bonus

Given the case you want to instantiate a class based on an interface you require it to have, it is more preformant to use is_a. There is only one exception - when the constructor is empty.

Example: is_a(<className>, <interfaceName>, true);

It will return bool. The third parameter "allow_string" allows it to check class names without instantiating the class.